We consider the system Applicative_first_order_05__#3.48. Alphabet: 0 : [] --> b 1 : [] --> b c : [b] --> b cons : [c * d] --> d f : [b] --> a false : [] --> a filter : [c -> a * d] --> d filter2 : [a * c -> a * c * d] --> d g : [b * b] --> b if : [a * b * b] --> b map : [c -> c * d] --> d nil : [] --> d s : [b] --> b true : [] --> a Rules: f(0) => true f(1) => false f(s(x)) => f(x) if(true, s(x), s(y)) => s(x) if(false, s(x), s(y)) => s(y) g(x, c(y)) => c(g(x, y)) g(x, c(y)) => g(x, if(f(x), c(g(s(x), y)), c(y))) map(h, nil) => nil map(h, cons(x, y)) => cons(h x, map(h, y)) filter(h, nil) => nil filter(h, cons(x, y)) => filter2(h x, h, x, y) filter2(true, h, x, y) => cons(x, filter(h, y)) filter2(false, h, x, y) => filter(h, 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] f#(s(X)) =#> f#(X) 1] g#(X, c(Y)) =#> g#(X, Y) 2] g#(X, c(Y)) =#> g#(X, if(f(X), c(g(s(X), Y)), c(Y))) 3] g#(X, c(Y)) =#> if#(f(X), c(g(s(X), Y)), c(Y)) 4] g#(X, c(Y)) =#> f#(X) 5] g#(X, c(Y)) =#> g#(s(X), Y) 6] map#(F, cons(X, Y)) =#> map#(F, Y) 7] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 8] filter2#(true, F, X, Y) =#> filter#(F, Y) 9] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: f(0) => true f(1) => false f(s(X)) => f(X) if(true, s(X), s(Y)) => s(X) if(false, s(X), s(Y)) => s(Y) g(X, c(Y)) => c(g(X, Y)) g(X, c(Y)) => g(X, if(f(X), c(g(s(X), Y)), c(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). The formative rules of (P_0, R_0) are R_1 ::= f(0) => true f(1) => false f(s(X)) => f(X) if(true, s(X), s(Y)) => s(X) if(false, s(X), s(Y)) => s(Y) g(X, c(Y)) => c(g(X, Y)) g(X, c(Y)) => g(X, if(f(X), c(g(s(X), Y)), c(Y))) map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, static, formative) by (P_0, R_1, static, formative). Thus, the original system is terminating if (P_0, R_1, static, formative) is finite. We consider the dependency pair problem (P_0, R_1, 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: f#(s(X)) >? f#(X) g#(X, c(Y)) >? g#(X, Y) g#(X, c(Y)) >? g#(X, if(f(X), c(g(s(X), Y)), c(Y))) g#(X, c(Y)) >? if#(f(X), c(g(s(X), Y)), c(Y)) g#(X, c(Y)) >? f#(X) g#(X, c(Y)) >? g#(s(X), 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) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, s(X), s(Y)) >= s(X) if(false, s(X), s(Y)) >= s(Y) g(X, c(Y)) >= c(g(X, Y)) g(X, c(Y)) >= g(X, if(f(X), c(g(s(X), Y)), c(Y))) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: 0 = 3 1 = 3 c = \y0.0 cons = \y0y1.0 f = \y0.2 f# = \y0.0 false = 0 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 g = \y0y1.0 g# = \y0y1.3 + 2y0 if = \y0y1y2.0 if# = \y0y1y2.0 map = \G0y1.2G0(0) map# = \G0y1.0 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] [[g#(_x0, c(_x1))]] = 3 + 2x0 >= 3 + 2x0 = [[g#(_x0, _x1)]] [[g#(_x0, c(_x1))]] = 3 + 2x0 >= 3 + 2x0 = [[g#(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[g#(_x0, c(_x1))]] = 3 + 2x0 > 0 = [[if#(f(_x0), c(g(s(_x0), _x1)), c(_x1))]] [[g#(_x0, c(_x1))]] = 3 + 2x0 > 0 = [[f#(_x0)]] [[g#(_x0, c(_x1))]] = 3 + 2x0 >= 3 = [[g#(s(_x0), _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)]] [[f(0)]] = 2 >= 0 = [[true]] [[f(1)]] = 2 >= 0 = [[false]] [[f(s(_x0))]] = 2 >= 2 = [[f(_x0)]] [[if(true, s(_x0), s(_x1))]] = 0 >= 0 = [[s(_x0)]] [[if(false, s(_x0), s(_x1))]] = 0 >= 0 = [[s(_x1)]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[c(g(_x0, _x1))]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[g(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[map(_F0, cons(_x1, _x2))]] = 2F0(0) >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[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_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of: f#(s(X)) =#> f#(X) g#(X, c(Y)) =#> g#(X, Y) g#(X, c(Y)) =#> g#(X, if(f(X), c(g(s(X), Y)), c(Y))) g#(X, c(Y)) =#> g#(s(X), 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_1, static, formative) is finite. We consider the dependency pair problem (P_1, R_1, 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: f#(s(X)) >? f#(X) g#(X, c(Y)) >? g#(X, Y) g#(X, c(Y)) >? g#(X, if(f(X), c(g(s(X), Y)), c(Y))) g#(X, c(Y)) >? g#(s(X), 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) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, s(X), s(Y)) >= s(X) if(false, s(X), s(Y)) >= s(Y) g(X, c(Y)) >= c(g(X, Y)) g(X, c(Y)) >= g(X, if(f(X), c(g(s(X), Y)), c(Y))) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: 0 = 3 1 = 3 c = \y0.0 cons = \y0y1.2 + y1 f = \y0.0 f# = \y0.0 false = 0 filter = \G0y1.y1 + G0(0) + G0(y1) + y1G0(y1) filter2 = \y0G1y2y3.2 + y3 + G1(0) + G1(y3) + y3G1(y3) filter2# = \y0G1y2y3.0 filter# = \G0y1.0 g = \y0y1.0 g# = \y0y1.2y0 if = \y0y1y2.2 map = \G0y1.2y1 map# = \G0y1.y1 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] [[g#(_x0, c(_x1))]] = 2x0 >= 2x0 = [[g#(_x0, _x1)]] [[g#(_x0, c(_x1))]] = 2x0 >= 2x0 = [[g#(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[g#(_x0, c(_x1))]] = 2x0 >= 0 = [[g#(s(_x0), _x1)]] [[map#(_F0, cons(_x1, _x2))]] = 2 + 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)]] [[f(0)]] = 0 >= 0 = [[true]] [[f(1)]] = 0 >= 0 = [[false]] [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] [[if(true, s(_x0), s(_x1))]] = 2 >= 0 = [[s(_x0)]] [[if(false, s(_x0), s(_x1))]] = 2 >= 0 = [[s(_x1)]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[c(g(_x0, _x1))]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[g(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 >= 2 + 2x2 = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x2 + F0(0) + 3F0(2 + x2) + x2F0(2 + x2) >= 2 + x2 + F0(0) + F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x2 + F0(0) + F0(x2) + x2F0(x2) >= 2 + x2 + F0(0) + F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x2 + F0(0) + F0(x2) + x2F0(x2) >= x2 + F0(0) + F0(x2) + x2F0(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_1, R_1, static, formative) by (P_2, R_1, static, formative), where P_2 consists of: f#(s(X)) =#> f#(X) g#(X, c(Y)) =#> g#(X, Y) g#(X, c(Y)) =#> g#(X, if(f(X), c(g(s(X), Y)), c(Y))) g#(X, c(Y)) =#> g#(s(X), 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_1, static, formative) is finite. We consider the dependency pair problem (P_2, R_1, 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: f#(s(X)) >? f#(X) g#(X, c(Y)) >? g#(X, Y) g#(X, c(Y)) >? g#(X, if(f(X), c(g(s(X), Y)), c(Y))) g#(X, c(Y)) >? g#(s(X), 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) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, s(X), s(Y)) >= s(X) if(false, s(X), s(Y)) >= s(Y) g(X, c(Y)) >= c(g(X, Y)) g(X, c(Y)) >= g(X, if(f(X), c(g(s(X), Y)), c(Y))) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: 0 = 3 1 = 3 c = \y0.2 + y0 cons = \y0y1.0 f = \y0.2 f# = \y0.0 false = 0 filter = \G0y1.0 filter2 = \y0G1y2y3.0 filter2# = \y0G1y2y3.0 filter# = \G0y1.0 g = \y0y1.y1 g# = \y0y1.y1 if = \y0y1y2.0 map = \G0y1.0 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] [[g#(_x0, c(_x1))]] = 2 + x1 > x1 = [[g#(_x0, _x1)]] [[g#(_x0, c(_x1))]] = 2 + x1 > 0 = [[g#(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[g#(_x0, c(_x1))]] = 2 + x1 > x1 = [[g#(s(_x0), _x1)]] [[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)]] [[f(0)]] = 2 >= 0 = [[true]] [[f(1)]] = 2 >= 0 = [[false]] [[f(s(_x0))]] = 2 >= 2 = [[f(_x0)]] [[if(true, s(_x0), s(_x1))]] = 0 >= 0 = [[s(_x0)]] [[if(false, s(_x0), s(_x1))]] = 0 >= 0 = [[s(_x1)]] [[g(_x0, c(_x1))]] = 2 + x1 >= 2 + x1 = [[c(g(_x0, _x1))]] [[g(_x0, c(_x1))]] = 2 + x1 >= 0 = [[g(_x0, if(f(_x0), c(g(s(_x0), _x1)), c(_x1)))]] [[map(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[cons(_F0 _x1, map(_F0, _x2))]] [[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_2, R_1, static, formative) by (P_3, R_1, static, formative), where P_3 consists of: f#(s(X)) =#> f#(X) 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_3, R_1, static, formative) is finite. We consider the dependency pair problem (P_3, R_1, static, formative). The formative rules of (P_3, R_1) are R_2 ::= f(0) => true f(1) => false f(s(X)) => f(X) if(true, s(X), s(Y)) => s(X) if(false, s(X), s(Y)) => s(Y) map(F, cons(X, Y)) => cons(F X, map(F, Y)) 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) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_1, static, formative) by (P_3, R_2, static, formative). Thus, the original system is terminating if (P_3, R_2, static, formative) is finite. We consider the dependency pair problem (P_3, R_2, 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: f#(s(X)) >? f#(X) 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) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, s(X), s(Y)) >= s(X) if(false, s(X), s(Y)) >= s(Y) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 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: 0 = 3 1 = 3 cons = \y0y1.3 + y0 + y1 f = \y0.3 f# = \y0.0 false = 2 filter = \G0y1.y1 filter2 = \y0G1y2y3.3 + y2 + y3 filter2# = \y0G1y2y3.2y0 + 3y3G1(y3) filter# = \G0y1.3 + 3y1G0(y1) if = \y0y1y2.3 map = \G0y1.y1 + 2y1G0(y1) s = \y0.0 true = 2 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 0 >= 0 = [[f#(_x0)]] [[filter#(_F0, cons(_x1, _x2))]] = 3 + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) > 2F0(x1) + 3x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 4 + 3x2F0(x2) > 3 + 3x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 4 + 3x2F0(x2) > 3 + 3x2F0(x2) = [[filter#(_F0, _x2)]] [[f(0)]] = 3 >= 2 = [[true]] [[f(1)]] = 3 >= 2 = [[false]] [[f(s(_x0))]] = 3 >= 3 = [[f(_x0)]] [[if(true, s(_x0), s(_x1))]] = 3 >= 0 = [[s(_x0)]] [[if(false, s(_x0), s(_x1))]] = 3 >= 0 = [[s(_x1)]] [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 6F0(3 + x1 + x2) >= 3 + x2 + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 >= 3 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 3 + x1 + x2 >= 3 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 3 + x1 + x2 >= 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_3, R_2, static, formative) by (P_4, R_2, static, formative), where P_4 consists of: f#(s(X)) =#> f#(X) Thus, the original system is terminating if (P_4, R_2, static, formative) is finite. We consider the dependency pair problem (P_4, R_2, static, formative). The formative rules of (P_4, R_2) are R_3 ::= f(0) => true f(1) => false f(s(X)) => f(X) if(true, s(X), s(Y)) => s(X) if(false, s(X), s(Y)) => s(Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_2, static, formative) by (P_4, R_3, static, formative). Thus, the original system is terminating if (P_4, R_3, static, formative) is finite. We consider the dependency pair problem (P_4, R_3, 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: f#(s(X)) >? f#(X) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, s(X), s(Y)) >= s(X) if(false, s(X), s(Y)) >= s(Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 1 = 3 f = \y0.3 + 2y0 f# = \y0.y0 false = 0 if = \y0y1y2.3 + 2y0 + 2y1 + 2y2 s = \y0.2 + 2y0 true = 2 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 2 + 2x0 > x0 = [[f#(_x0)]] [[f(0)]] = 9 >= 2 = [[true]] [[f(1)]] = 9 >= 0 = [[false]] [[f(s(_x0))]] = 7 + 4x0 >= 3 + 2x0 = [[f(_x0)]] [[if(true, s(_x0), s(_x1))]] = 15 + 4x0 + 4x1 >= 2 + 2x0 = [[s(_x0)]] [[if(false, s(_x0), s(_x1))]] = 11 + 4x0 + 4x1 >= 2 + 2x1 = [[s(_x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_3) by ({}, R_3). 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.