We consider the system Applicative_first_order_05__12. Alphabet: and : [a * a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d not : [a] --> a or : [a * a] --> a true : [] --> b Rules: not(not(x)) => x not(or(x, y)) => and(not(x), not(y)) not(and(x, y)) => or(not(x), not(y)) and(x, or(y, z)) => or(and(x, y), and(x, z)) and(or(x, y), z) => or(and(z, x), and(z, y)) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]). We thus obtain the following dependency pair problem (P_0, R_0, static, formative): Dependency Pairs P_0: 0] not#(or(X, Y)) =#> and#(not(X), not(Y)) 1] not#(or(X, Y)) =#> not#(X) 2] not#(or(X, Y)) =#> not#(Y) 3] not#(and(X, Y)) =#> not#(X) 4] not#(and(X, Y)) =#> not#(Y) 5] and#(X, or(Y, Z)) =#> and#(X, Y) 6] and#(X, or(Y, Z)) =#> and#(X, Z) 7] and#(or(X, Y), Z) =#> and#(Z, X) 8] and#(or(X, Y), Z) =#> and#(Z, Y) 9] map#(F, cons(X, Y)) =#> map#(F, Y) 10] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 11] filter2#(true, F, X, Y) =#> filter#(F, Y) 12] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: not(not(X)) => X not(or(X, Y)) => and(not(X), not(Y)) not(and(X, Y)) => or(not(X), not(Y)) and(X, or(Y, Z)) => or(and(X, Y), and(X, Z)) and(or(X, Y), Z) => or(and(Z, X), and(Z, Y)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. We consider the dependency pair problem (P_0, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? and#(not(X), not(Y)) not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.0 and# = \y0y1.0 cons = \y0y1.y1 false = 3 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 map = \G0y1.0 map# = \G0y1.0 nil = 0 not = \y0.2 + y0 not# = \y0.3 or = \y0y1.0 true = 3 Using this interpretation, the requirements translate to: [[not#(or(_x0, _x1))]] = 3 > 0 = [[and#(not(_x0), not(_x1))]] [[not#(or(_x0, _x1))]] = 3 >= 3 = [[not#(_x0)]] [[not#(or(_x0, _x1))]] = 3 >= 3 = [[not#(_x1)]] [[not#(and(_x0, _x1))]] = 3 >= 3 = [[not#(_x0)]] [[not#(and(_x0, _x1))]] = 3 >= 3 = [[not#(_x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[not(not(_x0))]] = 4 + x0 >= x0 = [[_x0]] [[not(or(_x0, _x1))]] = 2 >= 0 = [[and(not(_x0), not(_x1))]] [[not(and(_x0, _x1))]] = 2 >= 0 = [[or(not(_x0), not(_x1))]] [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 0 >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(X) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) and#(or(X, Y), Z) =#> and#(Z, Y) map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. We consider the dependency pair problem (P_1, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.0 and# = \y0y1.0 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 map = \G0y1.y1 map# = \G0y1.y1 nil = 0 not = \y0.2y0 not# = \y0.0 or = \y0y1.0 true = 3 Using this interpretation, the requirements translate to: [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 0 >= 0 = [[filter#(_F0, _x2)]] [[not(not(_x0))]] = 4x0 >= x0 = [[_x0]] [[not(or(_x0, _x1))]] = 0 >= 0 = [[and(not(_x0), not(_x1))]] [[not(and(_x0, _x1))]] = 0 >= 0 = [[or(not(_x0), not(_x1))]] [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(X) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) and#(or(X, Y), Z) =#> and#(Z, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_2, R_0, static, formative) is finite. We consider the dependency pair problem (P_2, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.0 and# = \y0y1.0 cons = \y0y1.1 + 2y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + 2y3 filter# = \G0y1.y1 map = \G0y1.y1 nil = 0 not = \y0.2y0 not# = \y0.0 or = \y0y1.0 true = 3 Using this interpretation, the requirements translate to: [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 > x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 > x2 = [[filter#(_F0, _x2)]] [[not(not(_x0))]] = 4x0 >= x0 = [[_x0]] [[not(or(_x0, _x1))]] = 0 >= 0 = [[and(not(_x0), not(_x1))]] [[not(and(_x0, _x1))]] = 0 >= 0 = [[or(not(_x0), not(_x1))]] [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] [[map(_F0, nil)]] = 0 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(X) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) and#(or(X, Y), Z) =#> and#(Z, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) Thus, the original system is terminating if (P_3, R_0, static, formative) is finite. We consider the dependency pair problem (P_3, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.0 and# = \y0y1.0 cons = \y0y1.0 false = 3 filter = \G0y1.2G0(0) filter2 = \y0G1y2y3.2G1(0) filter2# = \y0G1y2y3.0 filter# = \G0y1.3 map = \G0y1.2 nil = 0 not = \y0.2 + 2y0 not# = \y0.0 or = \y0y1.0 true = 3 Using this interpretation, the requirements translate to: [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(or(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x0)]] [[not#(and(_x0, _x1))]] = 0 >= 0 = [[not#(_x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x1)]] [[and#(_x0, or(_x1, _x2))]] = 0 >= 0 = [[and#(_x0, _x2)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x0)]] [[and#(or(_x0, _x1), _x2)]] = 0 >= 0 = [[and#(_x2, _x1)]] [[filter#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[not(not(_x0))]] = 6 + 4x0 >= x0 = [[_x0]] [[not(or(_x0, _x1))]] = 2 >= 0 = [[and(not(_x0), not(_x1))]] [[not(and(_x0, _x1))]] = 2 >= 0 = [[or(not(_x0), not(_x1))]] [[and(_x0, or(_x1, _x2))]] = 0 >= 0 = [[or(and(_x0, _x1), and(_x0, _x2))]] [[and(or(_x0, _x1), _x2)]] = 0 >= 0 = [[or(and(_x2, _x0), and(_x2, _x1))]] [[map(_F0, nil)]] = 2 >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 2F0(0) >= 2F0(0) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2F0(0) >= 0 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2F0(0) >= 2F0(0) = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(X) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) and#(or(X, Y), Z) =#> and#(Z, Y) Thus, the original system is terminating if (P_4, R_0, static, formative) is finite. We consider the dependency pair problem (P_4, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[and#(x_1, x_2)]] = and# [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_3, x_4) [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, and, and#, cons, false, filter, filter2, map, not, not#, or, true}, and the following precedence: false > filter = filter2 > map > @_{o -> o} > cons > and# > not > and > or > not# > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) >= not#(X) not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) > not#(X) not#(and(X, Y)) >= not#(Y) and# >= and# and# >= and# and# >= and# and# >= and# not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(F, X, Y) filter2(F, X, Y) >= cons(X, filter(F, Y)) filter2(F, X, Y) >= filter(F, Y) With these choices, we have: 1] not#(or(X, Y)) >= not#(X) because [2], by (Star) 2] not#*(or(X, Y)) >= not#(X) because not# in Mul and [3], by (Stat) 3] or(X, Y) > X because [4], by definition 4] or*(X, Y) >= X because [5], by (Select) 5] X >= X by (Meta) 6] not#(or(X, Y)) >= not#(Y) because [7], by (Star) 7] not#*(or(X, Y)) >= not#(Y) because [8], by (Select) 8] or(X, Y) >= not#(Y) because [9], by (Star) 9] or*(X, Y) >= not#(Y) because or > not# and [10], by (Copy) 10] or*(X, Y) >= Y because [11], by (Select) 11] Y >= Y by (Meta) 12] not#(and(X, Y)) > not#(X) because [13], by definition 13] not#*(and(X, Y)) >= not#(X) because not# in Mul and [14], by (Stat) 14] and(X, Y) > X because [15], by definition 15] and*(X, Y) >= X because [16], by (Select) 16] X >= X by (Meta) 17] not#(and(X, Y)) >= not#(Y) because [18], by (Star) 18] not#*(and(X, Y)) >= not#(Y) because [19], by (Select) 19] and(X, Y) >= not#(Y) because [20], by (Star) 20] and*(X, Y) >= not#(Y) because and > not# and [21], by (Copy) 21] and*(X, Y) >= Y because [22], by (Select) 22] Y >= Y by (Meta) 23] and# >= and# because and# in Mul, by (Fun) 24] and# >= and# because and# in Mul, by (Fun) 25] and# >= and# because and# in Mul, by (Fun) 26] and# >= and# because and# in Mul, by (Fun) 27] not(not(X)) >= X because [28], by (Star) 28] not*(not(X)) >= X because [29], by (Select) 29] not(X) >= X because [30], by (Star) 30] not*(X) >= X because [31], by (Select) 31] X >= X by (Meta) 32] not(or(X, Y)) >= and(not(X), not(Y)) because [33], by (Star) 33] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [34] and [35], by (Copy) 34] not*(or(X, Y)) >= not(X) because not in Mul and [3], by (Stat) 35] not*(or(X, Y)) >= not(Y) because not in Mul and [36], by (Stat) 36] or(X, Y) > Y because [10], by definition 37] not(and(X, Y)) >= or(not(X), not(Y)) because [38], by (Star) 38] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [39] and [40], by (Copy) 39] not*(and(X, Y)) >= not(X) because not in Mul and [14], by (Stat) 40] not*(and(X, Y)) >= not(Y) because not in Mul and [41], by (Stat) 41] and(X, Y) > Y because [21], by definition 42] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [43], by (Star) 43] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [44] and [49], by (Copy) 44] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [45] and [46], by (Stat) 45] X >= X by (Meta) 46] or(Y, Z) > Y because [47], by definition 47] or*(Y, Z) >= Y because [48], by (Select) 48] Y >= Y by (Meta) 49] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [45] and [50], by (Stat) 50] or(Y, Z) > Z because [51], by definition 51] or*(Y, Z) >= Z because [52], by (Select) 52] Z >= Z by (Meta) 53] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [54], by (Star) 54] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [55] and [60], by (Copy) 55] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [56] and [59], by (Stat) 56] or(X, Y) > X because [57], by definition 57] or*(X, Y) >= X because [58], by (Select) 58] X >= X by (Meta) 59] Z >= Z by (Meta) 60] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [61] and [59], by (Stat) 61] or(X, Y) > Y because [62], by definition 62] or*(X, Y) >= Y because [63], by (Select) 63] Y >= Y by (Meta) 64] map(F, _|_) >= _|_ by (Bot) 65] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [66], by (Star) 66] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [67] and [74], by (Copy) 67] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [68] and [70], by (Copy) 68] map*(F, cons(X, Y)) >= F because [69], by (Select) 69] F >= F by (Meta) 70] map*(F, cons(X, Y)) >= X because [71], by (Select) 71] cons(X, Y) >= X because [72], by (Star) 72] cons*(X, Y) >= X because [73], by (Select) 73] X >= X by (Meta) 74] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [75] and [76], by (Stat) 75] F >= F by (Meta) 76] cons(X, Y) > Y because [77], by definition 77] cons*(X, Y) >= Y because [78], by (Select) 78] Y >= Y by (Meta) 79] filter(F, _|_) >= _|_ by (Bot) 80] filter(F, cons(X, Y)) >= filter2(F, X, Y) because [81], by (Star) 81] filter*(F, cons(X, Y)) >= filter2(F, X, Y) because filter = filter2, filter in Mul, [82], [83] and [86], by (Stat) 82] F >= F by (Meta) 83] cons(X, Y) > X because [84], by definition 84] cons*(X, Y) >= X because [85], by (Select) 85] X >= X by (Meta) 86] cons(X, Y) > Y because [87], by definition 87] cons*(X, Y) >= Y because [88], by (Select) 88] Y >= Y by (Meta) 89] filter2(F, X, Y) >= cons(X, filter(F, Y)) because [90], by (Star) 90] filter2*(F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [91] and [93], by (Copy) 91] filter2*(F, X, Y) >= X because [92], by (Select) 92] X >= X by (Meta) 93] filter2*(F, X, Y) >= filter(F, Y) because filter2 = filter, filter2 in Mul, [94] and [95], by (Stat) 94] F >= F by (Meta) 95] Y >= Y by (Meta) 96] filter2(F, X, Y) >= filter(F, Y) because [97], by (Star) 97] filter2*(F, X, Y) >= filter(F, Y) because filter2 = filter, filter2 in Mul, [98] and [99], by (Stat) 98] F >= F by (Meta) 99] Y >= Y by (Meta) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) and#(or(X, Y), Z) =#> and#(Z, Y) Thus, the original system is terminating if (P_5, R_0, static, formative) is finite. We consider the dependency pair problem (P_5, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) and#(or(X, Y), Z) >? and#(Z, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[cons(x_1, x_2)]] = _|_ [[filter(x_1, x_2)]] = _|_ [[filter2(x_1, x_2, x_3, x_4)]] = _|_ [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, and, and#, false, map, not, not#, or, true}, and the following precedence: false > map > @_{o -> o} > and# > not > not# > and > or > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) >= not#(X) not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) >= not#(Y) and#(X, or(Y, Z)) >= and#(X, Y) and#(X, or(Y, Z)) >= and#(X, Z) and#(or(X, Y), Z) >= and#(Z, X) and#(or(X, Y), Z) > and#(Z, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, _|_) >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] not#(or(X, Y)) >= not#(X) because not# in Mul and [2], by (Fun) 2] or(X, Y) >= X because [3], by (Star) 3] or*(X, Y) >= X because [4], by (Select) 4] X >= X by (Meta) 5] not#(or(X, Y)) >= not#(Y) because [6], by (Star) 6] not#*(or(X, Y)) >= not#(Y) because not# in Mul and [7], by (Stat) 7] or(X, Y) > Y because [8], by definition 8] or*(X, Y) >= Y because [9], by (Select) 9] Y >= Y by (Meta) 10] not#(and(X, Y)) >= not#(Y) because not# in Mul and [11], by (Fun) 11] and(X, Y) >= Y because [12], by (Star) 12] and*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] and#(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [15] and [16], by (Fun) 15] X >= X by (Meta) 16] or(Y, Z) >= Y because [17], by (Star) 17] or*(Y, Z) >= Y because [18], by (Select) 18] Y >= Y by (Meta) 19] and#(X, or(Y, Z)) >= and#(X, Z) because [20], by (Star) 20] and#*(X, or(Y, Z)) >= and#(X, Z) because and# in Mul, [15] and [21], by (Stat) 21] or(Y, Z) > Z because [22], by definition 22] or*(Y, Z) >= Z because [23], by (Select) 23] Z >= Z by (Meta) 24] and#(or(X, Y), Z) >= and#(Z, X) because and# in Mul, [25] and [28], by (Fun) 25] or(X, Y) >= X because [26], by (Star) 26] or*(X, Y) >= X because [27], by (Select) 27] X >= X by (Meta) 28] Z >= Z by (Meta) 29] and#(or(X, Y), Z) > and#(Z, Y) because [30], by definition 30] and#*(or(X, Y), Z) >= and#(Z, Y) because and# in Mul, [31] and [28], by (Stat) 31] or(X, Y) > Y because [32], by definition 32] or*(X, Y) >= Y because [33], by (Select) 33] Y >= Y by (Meta) 34] not(not(X)) >= X because [35], by (Star) 35] not*(not(X)) >= X because [36], by (Select) 36] not(X) >= X because [37], by (Star) 37] not*(X) >= X because [38], by (Select) 38] X >= X by (Meta) 39] not(or(X, Y)) >= and(not(X), not(Y)) because [40], by (Star) 40] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [41] and [44], by (Copy) 41] not*(or(X, Y)) >= not(X) because not in Mul and [42], by (Stat) 42] or(X, Y) > X because [43], by definition 43] or*(X, Y) >= X because [4], by (Select) 44] not*(or(X, Y)) >= not(Y) because not in Mul and [7], by (Stat) 45] not(and(X, Y)) >= or(not(X), not(Y)) because [46], by (Star) 46] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [47] and [51], by (Copy) 47] not*(and(X, Y)) >= not(X) because not in Mul and [48], by (Stat) 48] and(X, Y) > X because [49], by definition 49] and*(X, Y) >= X because [50], by (Select) 50] X >= X by (Meta) 51] not*(and(X, Y)) >= not(Y) because not in Mul and [52], by (Stat) 52] and(X, Y) > Y because [53], by definition 53] and*(X, Y) >= Y because [13], by (Select) 54] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [55], by (Star) 55] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [56] and [59], by (Copy) 56] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [15] and [57], by (Stat) 57] or(Y, Z) > Y because [58], by definition 58] or*(Y, Z) >= Y because [18], by (Select) 59] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [15] and [21], by (Stat) 60] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [61], by (Star) 61] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [62] and [65], by (Copy) 62] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [63] and [28], by (Stat) 63] or(X, Y) > X because [64], by definition 64] or*(X, Y) >= X because [27], by (Select) 65] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [31] and [28], by (Stat) 66] map(F, _|_) >= _|_ by (Bot) 67] map(F, _|_) >= _|_ by (Bot) 68] _|_ >= _|_ by (Bot) 69] _|_ >= _|_ by (Bot) 70] _|_ >= _|_ by (Bot) 71] _|_ >= _|_ by (Bot) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_0, static, formative) by (P_6, R_0, static, formative), where P_6 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(X, or(Y, Z)) =#> and#(X, Z) and#(or(X, Y), Z) =#> and#(Z, X) Thus, the original system is terminating if (P_6, R_0, static, formative) is finite. We consider the dependency pair problem (P_6, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(X, or(Y, Z)) >? and#(X, Z) and#(or(X, Y), Z) >? and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[cons(x_1, x_2)]] = x_2 [[filter(x_1, x_2)]] = filter(x_2) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4) [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, and, and#, false, filter, filter2, map, not, not#, or, true}, and the following precedence: and# > false > @_{o -> o} > filter = filter2 > map > not > not# > and > or > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) >= not#(X) not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) >= not#(Y) and#(X, or(Y, Z)) >= and#(X, Y) and#(X, or(Y, Z)) > and#(X, Z) and#(or(X, Y), Z) >= and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, X) >= map(F, X) filter(_|_) >= _|_ filter(X) >= filter2(X) filter2(X) >= filter(X) filter2(X) >= filter(X) With these choices, we have: 1] not#(or(X, Y)) >= not#(X) because not# in Mul and [2], by (Fun) 2] or(X, Y) >= X because [3], by (Star) 3] or*(X, Y) >= X because [4], by (Select) 4] X >= X by (Meta) 5] not#(or(X, Y)) >= not#(Y) because not# in Mul and [6], by (Fun) 6] or(X, Y) >= Y because [7], by (Star) 7] or*(X, Y) >= Y because [8], by (Select) 8] Y >= Y by (Meta) 9] not#(and(X, Y)) >= not#(Y) because not# in Mul and [10], by (Fun) 10] and(X, Y) >= Y because [11], by (Star) 11] and*(X, Y) >= Y because [12], by (Select) 12] Y >= Y by (Meta) 13] and#(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [14] and [15], by (Fun) 14] X >= X by (Meta) 15] or(Y, Z) >= Y because [16], by (Star) 16] or*(Y, Z) >= Y because [17], by (Select) 17] Y >= Y by (Meta) 18] and#(X, or(Y, Z)) > and#(X, Z) because [19], by definition 19] and#*(X, or(Y, Z)) >= and#(X, Z) because and# in Mul, [14] and [20], by (Stat) 20] or(Y, Z) > Z because [21], by definition 21] or*(Y, Z) >= Z because [22], by (Select) 22] Z >= Z by (Meta) 23] and#(or(X, Y), Z) >= and#(Z, X) because and# in Mul, [24] and [27], by (Fun) 24] or(X, Y) >= X because [25], by (Star) 25] or*(X, Y) >= X because [26], by (Select) 26] X >= X by (Meta) 27] Z >= Z by (Meta) 28] not(not(X)) >= X because [29], by (Star) 29] not*(not(X)) >= X because [30], by (Select) 30] not(X) >= X because [31], by (Star) 31] not*(X) >= X because [32], by (Select) 32] X >= X by (Meta) 33] not(or(X, Y)) >= and(not(X), not(Y)) because [34], by (Star) 34] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [35] and [38], by (Copy) 35] not*(or(X, Y)) >= not(X) because not in Mul and [36], by (Stat) 36] or(X, Y) > X because [37], by definition 37] or*(X, Y) >= X because [4], by (Select) 38] not*(or(X, Y)) >= not(Y) because not in Mul and [39], by (Stat) 39] or(X, Y) > Y because [40], by definition 40] or*(X, Y) >= Y because [8], by (Select) 41] not(and(X, Y)) >= or(not(X), not(Y)) because [42], by (Star) 42] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [43] and [47], by (Copy) 43] not*(and(X, Y)) >= not(X) because not in Mul and [44], by (Stat) 44] and(X, Y) > X because [45], by definition 45] and*(X, Y) >= X because [46], by (Select) 46] X >= X by (Meta) 47] not*(and(X, Y)) >= not(Y) because not in Mul and [48], by (Stat) 48] and(X, Y) > Y because [49], by definition 49] and*(X, Y) >= Y because [12], by (Select) 50] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [51], by (Star) 51] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [52] and [55], by (Copy) 52] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [14] and [53], by (Stat) 53] or(Y, Z) > Y because [54], by definition 54] or*(Y, Z) >= Y because [17], by (Select) 55] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [14] and [20], by (Stat) 56] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [57], by (Star) 57] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [58] and [61], by (Copy) 58] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [59] and [27], by (Stat) 59] or(X, Y) > X because [60], by definition 60] or*(X, Y) >= X because [26], by (Select) 61] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [62] and [27], by (Stat) 62] or(X, Y) > Y because [63], by definition 63] or*(X, Y) >= Y because [64], by (Select) 64] Y >= Y by (Meta) 65] map(F, _|_) >= _|_ by (Bot) 66] map(F, X) >= map(F, X) because map in Mul, [67] and [68], by (Fun) 67] F >= F by (Meta) 68] X >= X by (Meta) 69] filter(_|_) >= _|_ by (Bot) 70] filter(X) >= filter2(X) because filter = filter2, filter in Mul and [71], by (Fun) 71] X >= X by (Meta) 72] filter2(X) >= filter(X) because filter2 = filter, filter2 in Mul and [73], by (Fun) 73] X >= X by (Meta) 74] filter2(X) >= filter(X) because filter2 = filter, filter2 in Mul and [75], by (Fun) 75] X >= X by (Meta) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_6, R_0, static, formative) by (P_7, R_0, static, formative), where P_7 consists of: not#(or(X, Y)) =#> not#(X) not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(or(X, Y), Z) =#> and#(Z, X) Thus, the original system is terminating if (P_7, R_0, static, formative) is finite. We consider the dependency pair problem (P_7, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(or(X, Y), Z) >? and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) [[nil]] = _|_ We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {and, and#, cons, false, map, not, not#, or, true}, and the following precedence: and# > false > map > @_{o -> o} = filter = filter2 > cons > not > and > or > not# > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) > not#(X) not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) >= not#(Y) and#(X, or(Y, Z)) >= and#(X, Y) and#(or(X, Y), Z) >= and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] not#(or(X, Y)) > not#(X) because [2], by definition 2] not#*(or(X, Y)) >= not#(X) because not# in Mul and [3], by (Stat) 3] or(X, Y) > X because [4], by definition 4] or*(X, Y) >= X because [5], by (Select) 5] X >= X by (Meta) 6] not#(or(X, Y)) >= not#(Y) because [7], by (Star) 7] not#*(or(X, Y)) >= not#(Y) because [8], by (Select) 8] or(X, Y) >= not#(Y) because [9], by (Star) 9] or*(X, Y) >= not#(Y) because or > not# and [10], by (Copy) 10] or*(X, Y) >= Y because [11], by (Select) 11] Y >= Y by (Meta) 12] not#(and(X, Y)) >= not#(Y) because [13], by (Star) 13] not#*(and(X, Y)) >= not#(Y) because not# in Mul and [14], by (Stat) 14] and(X, Y) > Y because [15], by definition 15] and*(X, Y) >= Y because [16], by (Select) 16] Y >= Y by (Meta) 17] and#(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [18] and [19], by (Fun) 18] X >= X by (Meta) 19] or(Y, Z) >= Y because [20], by (Star) 20] or*(Y, Z) >= Y because [21], by (Select) 21] Y >= Y by (Meta) 22] and#(or(X, Y), Z) >= and#(Z, X) because and# in Mul, [23] and [26], by (Fun) 23] or(X, Y) >= X because [24], by (Star) 24] or*(X, Y) >= X because [25], by (Select) 25] X >= X by (Meta) 26] Z >= Z by (Meta) 27] not(not(X)) >= X because [28], by (Star) 28] not*(not(X)) >= X because [29], by (Select) 29] not(X) >= X because [30], by (Star) 30] not*(X) >= X because [31], by (Select) 31] X >= X by (Meta) 32] not(or(X, Y)) >= and(not(X), not(Y)) because [33], by (Star) 33] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [34] and [35], by (Copy) 34] not*(or(X, Y)) >= not(X) because not in Mul and [3], by (Stat) 35] not*(or(X, Y)) >= not(Y) because not in Mul and [36], by (Stat) 36] or(X, Y) > Y because [10], by definition 37] not(and(X, Y)) >= or(not(X), not(Y)) because [38], by (Star) 38] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [39] and [43], by (Copy) 39] not*(and(X, Y)) >= not(X) because not in Mul and [40], by (Stat) 40] and(X, Y) > X because [41], by definition 41] and*(X, Y) >= X because [42], by (Select) 42] X >= X by (Meta) 43] not*(and(X, Y)) >= not(Y) because not in Mul and [14], by (Stat) 44] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [45], by (Star) 45] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [46] and [49], by (Copy) 46] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [18] and [47], by (Stat) 47] or(Y, Z) > Y because [48], by definition 48] or*(Y, Z) >= Y because [21], by (Select) 49] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [18] and [50], by (Stat) 50] or(Y, Z) > Z because [51], by definition 51] or*(Y, Z) >= Z because [52], by (Select) 52] Z >= Z by (Meta) 53] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [54], by (Star) 54] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [55] and [58], by (Copy) 55] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [56] and [26], by (Stat) 56] or(X, Y) > X because [57], by definition 57] or*(X, Y) >= X because [25], by (Select) 58] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [59] and [26], by (Stat) 59] or(X, Y) > Y because [60], by definition 60] or*(X, Y) >= Y because [61], by (Select) 61] Y >= Y by (Meta) 62] map(F, _|_) >= _|_ by (Bot) 63] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [64], by (Star) 64] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [65] and [72], by (Copy) 65] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [66] and [68], by (Copy) 66] map*(F, cons(X, Y)) >= F because [67], by (Select) 67] F >= F by (Meta) 68] map*(F, cons(X, Y)) >= X because [69], by (Select) 69] cons(X, Y) >= X because [70], by (Star) 70] cons*(X, Y) >= X because [71], by (Select) 71] X >= X by (Meta) 72] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [73] and [74], by (Stat) 73] F >= F by (Meta) 74] cons(X, Y) > Y because [75], by definition 75] cons*(X, Y) >= Y because [76], by (Select) 76] Y >= Y by (Meta) 77] filter(F, _|_) >= _|_ by (Bot) 78] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [79], by (Star) 79] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [80], [81], [84], [88], [89] and [91], by (Stat) 80] F >= F by (Meta) 81] cons(X, Y) > Y because [82], by definition 82] cons*(X, Y) >= Y because [83], by (Select) 83] Y >= Y by (Meta) 84] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter = @_{o -> o}, [80], [85], [88] and [89], by (Stat) 85] cons(X, Y) > X because [86], by definition 86] cons*(X, Y) >= X because [87], by (Select) 87] X >= X by (Meta) 88] filter*(F, cons(X, Y)) >= F because [80], by (Select) 89] filter*(F, cons(X, Y)) >= X because [90], by (Select) 90] cons(X, Y) >= X because [86], by (Star) 91] filter*(F, cons(X, Y)) >= Y because [92], by (Select) 92] cons(X, Y) >= Y because [82], by (Star) 93] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [94], by (Star) 94] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [95] and [97], by (Copy) 95] filter2*(true, F, X, Y) >= X because [96], by (Select) 96] X >= X by (Meta) 97] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [98], [99], [100] and [101], by (Stat) 98] F >= F by (Meta) 99] Y >= Y by (Meta) 100] filter2*(true, F, X, Y) >= F because [98], by (Select) 101] filter2*(true, F, X, Y) >= Y because [99], by (Select) 102] filter2(false, F, X, Y) >= filter(F, Y) because [103], by (Star) 103] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [104], [105], [106] and [107], by (Stat) 104] F >= F by (Meta) 105] Y >= Y by (Meta) 106] filter2*(false, F, X, Y) >= F because [104], by (Select) 107] filter2*(false, F, X, Y) >= Y because [105], by (Select) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_7, R_0, static, formative) by (P_8, R_0, static, formative), where P_8 consists of: not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) and#(or(X, Y), Z) =#> and#(Z, X) Thus, the original system is terminating if (P_8, R_0, static, formative) is finite. We consider the dependency pair problem (P_8, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) and#(or(X, Y), Z) >? and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) [[nil]] = _|_ We choose Lex = {filter, filter2} and Mul = {@_{o -> o}, and, and#, cons, false, map, not, not#, or, true}, and the following precedence: and# > false > filter = filter2 > map > @_{o -> o} > cons > not > not# > and > or > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) >= not#(Y) and#(X, or(Y, Z)) >= and#(X, Y) and#(or(X, Y), Z) > and#(Z, X) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] not#(or(X, Y)) >= not#(Y) because not# in Mul and [2], by (Fun) 2] or(X, Y) >= Y because [3], by (Star) 3] or*(X, Y) >= Y because [4], by (Select) 4] Y >= Y by (Meta) 5] not#(and(X, Y)) >= not#(Y) because [6], by (Star) 6] not#*(and(X, Y)) >= not#(Y) because not# in Mul and [7], by (Stat) 7] and(X, Y) > Y because [8], by definition 8] and*(X, Y) >= Y because [9], by (Select) 9] Y >= Y by (Meta) 10] and#(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [11] and [12], by (Fun) 11] X >= X by (Meta) 12] or(Y, Z) >= Y because [13], by (Star) 13] or*(Y, Z) >= Y because [14], by (Select) 14] Y >= Y by (Meta) 15] and#(or(X, Y), Z) > and#(Z, X) because [16], by definition 16] and#*(or(X, Y), Z) >= and#(Z, X) because and# in Mul, [17] and [20], by (Stat) 17] or(X, Y) > X because [18], by definition 18] or*(X, Y) >= X because [19], by (Select) 19] X >= X by (Meta) 20] Z >= Z by (Meta) 21] not(not(X)) >= X because [22], by (Star) 22] not*(not(X)) >= X because [23], by (Select) 23] not(X) >= X because [24], by (Star) 24] not*(X) >= X because [25], by (Select) 25] X >= X by (Meta) 26] not(or(X, Y)) >= and(not(X), not(Y)) because [27], by (Star) 27] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [28] and [32], by (Copy) 28] not*(or(X, Y)) >= not(X) because not in Mul and [29], by (Stat) 29] or(X, Y) > X because [30], by definition 30] or*(X, Y) >= X because [31], by (Select) 31] X >= X by (Meta) 32] not*(or(X, Y)) >= not(Y) because not in Mul and [33], by (Stat) 33] or(X, Y) > Y because [34], by definition 34] or*(X, Y) >= Y because [4], by (Select) 35] not(and(X, Y)) >= or(not(X), not(Y)) because [36], by (Star) 36] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [37] and [41], by (Copy) 37] not*(and(X, Y)) >= not(X) because not in Mul and [38], by (Stat) 38] and(X, Y) > X because [39], by definition 39] and*(X, Y) >= X because [40], by (Select) 40] X >= X by (Meta) 41] not*(and(X, Y)) >= not(Y) because not in Mul and [7], by (Stat) 42] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [43], by (Star) 43] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [44] and [47], by (Copy) 44] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [11] and [45], by (Stat) 45] or(Y, Z) > Y because [46], by definition 46] or*(Y, Z) >= Y because [14], by (Select) 47] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [11] and [48], by (Stat) 48] or(Y, Z) > Z because [49], by definition 49] or*(Y, Z) >= Z because [50], by (Select) 50] Z >= Z by (Meta) 51] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [52], by (Star) 52] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [53] and [54], by (Copy) 53] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [17] and [20], by (Stat) 54] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [55] and [20], by (Stat) 55] or(X, Y) > Y because [56], by definition 56] or*(X, Y) >= Y because [57], by (Select) 57] Y >= Y by (Meta) 58] map(F, _|_) >= _|_ by (Bot) 59] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [60], by (Star) 60] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [61] and [68], by (Copy) 61] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [62] and [64], by (Copy) 62] map*(F, cons(X, Y)) >= F because [63], by (Select) 63] F >= F by (Meta) 64] map*(F, cons(X, Y)) >= X because [65], by (Select) 65] cons(X, Y) >= X because [66], by (Star) 66] cons*(X, Y) >= X because [67], by (Select) 67] X >= X by (Meta) 68] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [69] and [70], by (Stat) 69] F >= F by (Meta) 70] cons(X, Y) > Y because [71], by definition 71] cons*(X, Y) >= Y because [72], by (Select) 72] Y >= Y by (Meta) 73] filter(F, _|_) >= _|_ by (Bot) 74] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [75], by (Star) 75] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [76], [77], [80], [81], [82] and [86], by (Stat) 76] F >= F by (Meta) 77] cons(X, Y) > Y because [78], by definition 78] cons*(X, Y) >= Y because [79], by (Select) 79] Y >= Y by (Meta) 80] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [81] and [82], by (Copy) 81] filter*(F, cons(X, Y)) >= F because [76], by (Select) 82] filter*(F, cons(X, Y)) >= X because [83], by (Select) 83] cons(X, Y) >= X because [84], by (Star) 84] cons*(X, Y) >= X because [85], by (Select) 85] X >= X by (Meta) 86] filter*(F, cons(X, Y)) >= Y because [87], by (Select) 87] cons(X, Y) >= Y because [78], by (Star) 88] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [89], by (Star) 89] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [90] and [92], by (Copy) 90] filter2*(true, F, X, Y) >= X because [91], by (Select) 91] X >= X by (Meta) 92] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [93], [94], [95] and [96], by (Stat) 93] F >= F by (Meta) 94] Y >= Y by (Meta) 95] filter2*(true, F, X, Y) >= F because [93], by (Select) 96] filter2*(true, F, X, Y) >= Y because [94], by (Select) 97] filter2(false, F, X, Y) >= filter(F, Y) because [98], by (Star) 98] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [99], [100], [101] and [102], by (Stat) 99] F >= F by (Meta) 100] Y >= Y by (Meta) 101] filter2*(false, F, X, Y) >= F because [99], by (Select) 102] filter2*(false, F, X, Y) >= Y because [100], by (Select) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_8, R_0, static, formative) by (P_9, R_0, static, formative), where P_9 consists of: not#(or(X, Y)) =#> not#(Y) not#(and(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) Thus, the original system is terminating if (P_9, R_0, static, formative) is finite. We consider the dependency pair problem (P_9, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_2, x_4, x_1, x_3) [[nil]] = _|_ We choose Lex = {filter, filter2} and Mul = {@_{o -> o}, and, and#, cons, false, map, not, not#, or, true}, and the following precedence: and# > false > filter = filter2 > map > @_{o -> o} > cons > not > and = not# > or > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: not#(or(X, Y)) >= not#(Y) not#(and(X, Y)) > not#(Y) and#(X, or(Y, Z)) >= and#(X, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] not#(or(X, Y)) >= not#(Y) because not# in Mul and [2], by (Fun) 2] or(X, Y) >= Y because [3], by (Star) 3] or*(X, Y) >= Y because [4], by (Select) 4] Y >= Y by (Meta) 5] not#(and(X, Y)) > not#(Y) because [6], by definition 6] not#*(and(X, Y)) >= not#(Y) because [7], by (Select) 7] and(X, Y) >= not#(Y) because [8], by (Star) 8] and*(X, Y) >= not#(Y) because and = not#, and in Mul and [9], by (Stat) 9] Y >= Y by (Meta) 10] and#(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [11] and [12], by (Fun) 11] X >= X by (Meta) 12] or(Y, Z) >= Y because [13], by (Star) 13] or*(Y, Z) >= Y because [14], by (Select) 14] Y >= Y by (Meta) 15] not(not(X)) >= X because [16], by (Star) 16] not*(not(X)) >= X because [17], by (Select) 17] not(X) >= X because [18], by (Star) 18] not*(X) >= X because [19], by (Select) 19] X >= X by (Meta) 20] not(or(X, Y)) >= and(not(X), not(Y)) because [21], by (Star) 21] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [22] and [26], by (Copy) 22] not*(or(X, Y)) >= not(X) because not in Mul and [23], by (Stat) 23] or(X, Y) > X because [24], by definition 24] or*(X, Y) >= X because [25], by (Select) 25] X >= X by (Meta) 26] not*(or(X, Y)) >= not(Y) because not in Mul and [27], by (Stat) 27] or(X, Y) > Y because [28], by definition 28] or*(X, Y) >= Y because [4], by (Select) 29] not(and(X, Y)) >= or(not(X), not(Y)) because [30], by (Star) 30] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [31] and [35], by (Copy) 31] not*(and(X, Y)) >= not(X) because not in Mul and [32], by (Stat) 32] and(X, Y) > X because [33], by definition 33] and*(X, Y) >= X because [34], by (Select) 34] X >= X by (Meta) 35] not*(and(X, Y)) >= not(Y) because not in Mul and [36], by (Stat) 36] and(X, Y) > Y because [37], by definition 37] and*(X, Y) >= Y because [9], by (Select) 38] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [39], by (Star) 39] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [40] and [43], by (Copy) 40] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [11] and [41], by (Stat) 41] or(Y, Z) > Y because [42], by definition 42] or*(Y, Z) >= Y because [14], by (Select) 43] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [11] and [44], by (Stat) 44] or(Y, Z) > Z because [45], by definition 45] or*(Y, Z) >= Z because [46], by (Select) 46] Z >= Z by (Meta) 47] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [48], by (Star) 48] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [49] and [54], by (Copy) 49] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [50] and [53], by (Stat) 50] or(X, Y) > X because [51], by definition 51] or*(X, Y) >= X because [52], by (Select) 52] X >= X by (Meta) 53] Z >= Z by (Meta) 54] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [55] and [53], by (Stat) 55] or(X, Y) > Y because [56], by definition 56] or*(X, Y) >= Y because [57], by (Select) 57] Y >= Y by (Meta) 58] map(F, _|_) >= _|_ by (Bot) 59] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [60], by (Star) 60] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [61] and [68], by (Copy) 61] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map > @_{o -> o}, [62] and [64], by (Copy) 62] map*(F, cons(X, Y)) >= F because [63], by (Select) 63] F >= F by (Meta) 64] map*(F, cons(X, Y)) >= X because [65], by (Select) 65] cons(X, Y) >= X because [66], by (Star) 66] cons*(X, Y) >= X because [67], by (Select) 67] X >= X by (Meta) 68] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [69] and [70], by (Stat) 69] F >= F by (Meta) 70] cons(X, Y) > Y because [71], by definition 71] cons*(X, Y) >= Y because [72], by (Select) 72] Y >= Y by (Meta) 73] filter(F, _|_) >= _|_ by (Bot) 74] filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because [75], by (Star) 75] filter*(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) because filter = filter2, [76], [77], [80], [81], [82] and [86], by (Stat) 76] F >= F by (Meta) 77] cons(X, Y) > Y because [78], by definition 78] cons*(X, Y) >= Y because [79], by (Select) 79] Y >= Y by (Meta) 80] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [81] and [82], by (Copy) 81] filter*(F, cons(X, Y)) >= F because [76], by (Select) 82] filter*(F, cons(X, Y)) >= X because [83], by (Select) 83] cons(X, Y) >= X because [84], by (Star) 84] cons*(X, Y) >= X because [85], by (Select) 85] X >= X by (Meta) 86] filter*(F, cons(X, Y)) >= Y because [87], by (Select) 87] cons(X, Y) >= Y because [78], by (Star) 88] filter2(true, F, X, Y) >= cons(X, filter(F, Y)) because [89], by (Star) 89] filter2*(true, F, X, Y) >= cons(X, filter(F, Y)) because filter2 > cons, [90] and [92], by (Copy) 90] filter2*(true, F, X, Y) >= X because [91], by (Select) 91] X >= X by (Meta) 92] filter2*(true, F, X, Y) >= filter(F, Y) because filter2 = filter, [93], [94], [95] and [96], by (Stat) 93] F >= F by (Meta) 94] Y >= Y by (Meta) 95] filter2*(true, F, X, Y) >= F because [93], by (Select) 96] filter2*(true, F, X, Y) >= Y because [94], by (Select) 97] filter2(false, F, X, Y) >= filter(F, Y) because [98], by (Star) 98] filter2*(false, F, X, Y) >= filter(F, Y) because filter2 = filter, [99], [100], [101] and [102], by (Stat) 99] F >= F by (Meta) 100] Y >= Y by (Meta) 101] filter2*(false, F, X, Y) >= F because [99], by (Select) 102] filter2*(false, F, X, Y) >= Y because [100], by (Select) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_9, R_0, static, formative) by (P_10, R_0, static, formative), where P_10 consists of: not#(or(X, Y)) =#> not#(Y) and#(X, or(Y, Z)) =#> and#(X, Y) Thus, the original system is terminating if (P_10, R_0, static, formative) is finite. We consider the dependency pair problem (P_10, R_0, static, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: not#(or(X, Y)) >? not#(Y) and#(X, or(Y, Z)) >? and#(X, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, nil) >= nil map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, nil) >= nil filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[cons(x_1, x_2)]] = x_2 [[filter(x_1, x_2)]] = _|_ [[filter2(x_1, x_2, x_3, x_4)]] = _|_ [[nil]] = _|_ [[not#(x_1)]] = x_1 We choose Lex = {} and Mul = {@_{o -> o}, and, and#, false, map, not, or, true}, and the following precedence: @_{o -> o} > and# > false > map > not > and > or > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: or(X, Y) > Y and#(X, or(Y, Z)) > and#(X, Y) not(not(X)) >= X not(or(X, Y)) >= and(not(X), not(Y)) not(and(X, Y)) >= or(not(X), not(Y)) and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) map(F, _|_) >= _|_ map(F, X) >= map(F, X) _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ _|_ >= _|_ With these choices, we have: 1] or(X, Y) > Y because [2], by definition 2] or*(X, Y) >= Y because [3], by (Select) 3] Y >= Y by (Meta) 4] and#(X, or(Y, Z)) > and#(X, Y) because [5], by definition 5] and#*(X, or(Y, Z)) >= and#(X, Y) because and# in Mul, [6] and [7], by (Stat) 6] X >= X by (Meta) 7] or(Y, Z) > Y because [8], by definition 8] or*(Y, Z) >= Y because [9], by (Select) 9] Y >= Y by (Meta) 10] not(not(X)) >= X because [11], by (Star) 11] not*(not(X)) >= X because [12], by (Select) 12] not(X) >= X because [13], by (Star) 13] not*(X) >= X because [14], by (Select) 14] X >= X by (Meta) 15] not(or(X, Y)) >= and(not(X), not(Y)) because [16], by (Star) 16] not*(or(X, Y)) >= and(not(X), not(Y)) because not > and, [17] and [21], by (Copy) 17] not*(or(X, Y)) >= not(X) because not in Mul and [18], by (Stat) 18] or(X, Y) > X because [19], by definition 19] or*(X, Y) >= X because [20], by (Select) 20] X >= X by (Meta) 21] not*(or(X, Y)) >= not(Y) because not in Mul and [22], by (Stat) 22] or(X, Y) > Y because [2], by definition 23] not(and(X, Y)) >= or(not(X), not(Y)) because [24], by (Star) 24] not*(and(X, Y)) >= or(not(X), not(Y)) because not > or, [25] and [29], by (Copy) 25] not*(and(X, Y)) >= not(X) because not in Mul and [26], by (Stat) 26] and(X, Y) > X because [27], by definition 27] and*(X, Y) >= X because [28], by (Select) 28] X >= X by (Meta) 29] not*(and(X, Y)) >= not(Y) because not in Mul and [30], by (Stat) 30] and(X, Y) > Y because [31], by definition 31] and*(X, Y) >= Y because [32], by (Select) 32] Y >= Y by (Meta) 33] and(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because [34], by (Star) 34] and*(X, or(Y, Z)) >= or(and(X, Y), and(X, Z)) because and > or, [35] and [36], by (Copy) 35] and*(X, or(Y, Z)) >= and(X, Y) because and in Mul, [6] and [7], by (Stat) 36] and*(X, or(Y, Z)) >= and(X, Z) because and in Mul, [6] and [37], by (Stat) 37] or(Y, Z) > Z because [38], by definition 38] or*(Y, Z) >= Z because [39], by (Select) 39] Z >= Z by (Meta) 40] and(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because [41], by (Star) 41] and*(or(X, Y), Z) >= or(and(Z, X), and(Z, Y)) because and > or, [42] and [47], by (Copy) 42] and*(or(X, Y), Z) >= and(Z, X) because and in Mul, [43] and [46], by (Stat) 43] or(X, Y) > X because [44], by definition 44] or*(X, Y) >= X because [45], by (Select) 45] X >= X by (Meta) 46] Z >= Z by (Meta) 47] and*(or(X, Y), Z) >= and(Z, Y) because and in Mul, [48] and [46], by (Stat) 48] or(X, Y) > Y because [49], by definition 49] or*(X, Y) >= Y because [50], by (Select) 50] Y >= Y by (Meta) 51] map(F, _|_) >= _|_ by (Bot) 52] map(F, X) >= map(F, X) because map in Mul, [53] and [54], by (Fun) 53] F >= F by (Meta) 54] X >= X by (Meta) 55] _|_ >= _|_ by (Bot) 56] _|_ >= _|_ by (Bot) 57] _|_ >= _|_ by (Bot) 58] _|_ >= _|_ by (Bot) By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_10, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.