We consider the system Applicative_first_order_05__perfect2. Alphabet: 0 : [] --> a cons : [c * d] --> d f : [a * a * a * a] --> b false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d if : [b * b * b] --> b le : [a * a] --> b map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d perfectp : [a] --> b s : [a] --> a true : [] --> b Rules: minus(0, x) => 0 minus(s(x), 0) => s(x) minus(s(x), s(y)) => minus(x, y) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) if(true, x, y) => x if(false, x, y) => y perfectp(0) => false perfectp(s(x)) => f(x, s(0), s(x), s(x)) f(0, x, 0, y) => true f(0, x, s(y), z) => false f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, 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 dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] le#(s(X), s(Y)) =#> le#(X, Y) 2] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 3] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 4] f#(s(X), 0, Y, Z) =#> minus#(Y, s(X)) 5] f#(s(X), s(Y), Z, U) =#> if#(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 6] f#(s(X), s(Y), Z, U) =#> le#(X, Y) 7] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 8] f#(s(X), s(Y), Z, U) =#> minus#(Y, X) 9] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) 10] map#(F, cons(X, Y)) =#> F(X) 11] map#(F, cons(X, Y)) =#> map#(F, Y) 12] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 13] filter#(F, cons(X, Y)) =#> F(X) 14] filter2#(true, F, X, Y) =#> filter#(F, Y) 15] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 1 * 2 : 5, 6, 7, 8, 9 * 3 : 3, 4, 5, 6, 7, 8, 9 * 4 : 0 * 5 : * 6 : 1 * 7 : 3, 4, 5, 6, 7, 8, 9 * 8 : 0 * 9 : 3, 4, 5, 6, 7, 8, 9 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 * 11 : 10, 11 * 12 : 14, 15 * 13 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 * 14 : 12, 13 * 15 : 12, 13 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) P_4: map#(F, cons(X, Y)) =#> F(X) map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). The formative rules of (P_4, R_0) are R_1 ::= minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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_4, R_0, minimal, formative) by (P_4, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: map#(F, cons(X, Y)) >? F(X) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) minus(0, X) >= 0 minus(s(X), 0) >= s(X) minus(s(X), s(Y)) >= minus(X, Y) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) if(true, X, Y) >= X if(false, X, Y) >= Y perfectp(0) >= false perfectp(s(X)) >= f(X, s(0), s(X), s(X)) f(0, X, 0, Y) >= true f(0, X, s(Y), Z) >= false f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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 = 0 cons = \y0y1.2 + y1 + 2y0 f = \y0y1y2y3.0 false = 0 filter = \G0y1.2y1 + G0(y1) + 2y1G0(y1) filter2 = \y0G1y2y3.2 + 2y0 + 2y2 + 2y3 + G1(y3) + 2y3G1(y3) filter2# = \y0G1y2y3.G1(0) + y3G1(y3) filter# = \G0y1.G0(0) + y1G0(y1) if = \y0y1y2.y1 + y2 le = \y0y1.3 map = \G0y1.2y1 + 2y1G0(y1) + 2G0(y1) map# = \G0y1.3 + y1 + G0(y1) + y1G0(y1) minus = \y0y1.3 + 2y0 + 2y1 perfectp = \y0.3 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 3F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + 2x1F0(2 + x2 + 2x1) + 3F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 3 + x2 + F0(x2) + x2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= F0(0) + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = F0(0) + x2F0(x2) >= F0(0) + x2F0(x2) = [[filter#(_F0, _x2)]] [[minus(0, _x0)]] = 3 + 2x0 >= 0 = [[0]] [[minus(s(_x0), 0)]] = 3 + 4x0 >= 2x0 = [[s(_x0)]] [[minus(s(_x0), s(_x1))]] = 3 + 4x0 + 4x1 >= 3 + 2x0 + 2x1 = [[minus(_x0, _x1)]] [[le(0, _x0)]] = 3 >= 0 = [[true]] [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[perfectp(0)]] = 3 >= 0 = [[false]] [[perfectp(s(_x0))]] = 3 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 6F0(2 + x2 + 2x1) >= 2 + 2x2 + 2x2F0(x2) + 2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 5F0(2 + x2 + 2x1) >= 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) + 2max(x1, F0(x1)) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) >= 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + 2x1 + 2x2 + F0(x2) + 2x2F0(x2) >= 2x2 + F0(x2) + 2x2F0(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_4, R_1, minimal, formative) by (P_5, R_1, minimal, formative), where P_5 consists of: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) minus(0, X) >= 0 minus(s(X), 0) >= s(X) minus(s(X), s(Y)) >= minus(X, Y) le(0, X) >= true le(s(X), 0) >= false le(s(X), s(Y)) >= le(X, Y) if(true, X, Y) >= X if(false, X, Y) >= Y perfectp(0) >= false perfectp(s(X)) >= f(X, s(0), s(X), s(X)) f(0, X, 0, Y) >= true f(0, X, s(Y), Z) >= false f(s(X), 0, Y, Z) >= f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) >= if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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 = 1 cons = \y0y1.1 + y0 + y1 f = \y0y1y2y3.0 false = 0 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + y2 + y3 filter2# = \y0G1y2y3.1 + y3G1(y3) filter# = \G0y1.1 + y1G0(y1) if = \y0y1y2.y1 + y2 le = \y0y1.3 map = \G0y1.y1 + 2y1G0(y1) minus = \y0y1.3 + 2y0 + 2y1 perfectp = \y0.3 + y0 s = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) >= 1 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 1 + F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2F0(x2) >= 1 + x2F0(x2) = [[filter#(_F0, _x2)]] [[minus(0, _x0)]] = 5 + 2x0 >= 1 = [[0]] [[minus(s(_x0), 0)]] = 5 + 4x0 >= 2x0 = [[s(_x0)]] [[minus(s(_x0), s(_x1))]] = 3 + 4x0 + 4x1 >= 3 + 2x0 + 2x1 = [[minus(_x0, _x1)]] [[le(0, _x0)]] = 3 >= 0 = [[true]] [[le(s(_x0), 0)]] = 3 >= 0 = [[false]] [[le(s(_x0), s(_x1))]] = 3 >= 3 = [[le(_x0, _x1)]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[perfectp(0)]] = 4 >= 0 = [[false]] [[perfectp(s(_x0))]] = 3 + 2x0 >= 0 = [[f(_x0, s(0), s(_x0), s(_x0))]] [[f(0, _x0, 0, _x1)]] = 0 >= 0 = [[true]] [[f(0, _x0, s(_x1), _x2)]] = 0 >= 0 = [[false]] [[f(s(_x0), 0, _x1, _x2)]] = 0 >= 0 = [[f(_x0, _x2, minus(_x1, s(_x0)), _x2)]] [[f(s(_x0), s(_x1), _x2, _x3)]] = 0 >= 0 = [[if(le(_x0, _x1), f(s(_x0), minus(_x1, _x0), _x2, _x3), f(_x0, _x3, _x2, _x3))]] [[map(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) >= 1 + x2 + 2x2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 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_5, R_1, minimal, formative) by (P_6, R_1, minimal, formative), where P_6 consists of: 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 each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_6, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_1, minimal, formative). We apply the subterm criterion with the following projection function: nu(filter2#) = 4 nu(filter#) = 2 Thus, we can orient the dependency pairs as follows: nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, R_1, minimal, f) by (P_7, R_1, minimal, f), where P_7 contains: filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_7, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_1, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), 0, Y, Z)) = s(X) |> X = nu(f#(X, Z, minus(Y, s(X)), Z)) nu(f#(s(X), s(Y), Z, U)) = s(X) = s(X) = nu(f#(s(X), minus(Y, X), Z, U)) nu(f#(s(X), s(Y), Z, U)) = s(X) |> X = nu(f#(X, U, Z, U)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_8, R_0, minimal, f), where P_8 contains: f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_8, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_8, R_0, minimal, formative). The formative rules of (P_8, R_0) are R_2 ::= minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_8, R_0, minimal, formative) by (P_8, R_2, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_8, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_8, R_2, minimal, 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), s(Y), Z, U) >? f#(s(X), minus(Y, X), Z, U) minus(0, X) >= 0 minus(s(X), 0) >= s(X) minus(s(X), s(Y)) >= minus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 f# = \y0y1y2y3.y1 minus = \y0y1.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[f#(s(_x0), s(_x1), _x2, _x3)]] = 1 + 2x1 > x1 = [[f#(s(_x0), minus(_x1, _x0), _x2, _x3)]] [[minus(0, _x0)]] = 0 >= 0 = [[0]] [[minus(s(_x0), 0)]] = 1 + 2x0 >= 1 + 2x0 = [[s(_x0)]] [[minus(s(_x0), s(_x1))]] = 1 + 2x0 >= x0 = [[minus(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_8, R_2) by ({}, R_2). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(le#) = 1 Thus, we can orient the dependency pairs as follows: nu(le#(s(X), s(Y))) = s(X) |> X = nu(le#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). 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.