We consider the system Applicative_first_order_05__#3.22. Alphabet: 0 : [] --> 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 plus : [a * a] --> a s : [a] --> a times : [a * a] --> a true : [] --> b Rules: times(x, plus(y, s(z))) => plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) times(x, 0) => 0 times(x, s(y)) => plus(times(x, y), x) plus(x, 0) => x plus(x, s(y)) => s(plus(x, 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] times#(X, plus(Y, s(Z))) =#> plus#(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 1] times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) 2] times#(X, plus(Y, s(Z))) =#> plus#(Y, times(s(Z), 0)) 3] times#(X, plus(Y, s(Z))) =#> times#(s(Z), 0) 4] times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) 5] times#(X, s(Y)) =#> plus#(times(X, Y), X) 6] times#(X, s(Y)) =#> times#(X, Y) 7] plus#(X, s(Y)) =#> plus#(X, Y) 8] map#(F, cons(X, Y)) =#> map#(F, Y) 9] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 10] filter2#(true, F, X, Y) =#> filter#(F, Y) 11] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(X, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 7 * 1 : 0, 1, 2, 3, 4, 5, 6 * 2 : 7 * 3 : * 4 : 5, 6 * 5 : 7 * 6 : 0, 1, 2, 3, 4, 5, 6 * 7 : 7 * 8 : 8 * 9 : 10, 11 * 10 : 9 * 11 : 9 This graph has the following strongly connected components: P_1: times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) times#(X, s(Y)) =#> times#(X, Y) P_2: plus#(X, s(Y)) =#> plus#(X, Y) P_3: map#(F, cons(X, Y)) =#> map#(F, Y) P_4: 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) 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, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (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: 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) times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, 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: 0 = 0 cons = \y0y1.1 + y1 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + y3 filter# = \G0y1.y1 map = \G0y1.y1 nil = 0 plus = \y0y1.y0 s = \y0.0 times = \y0y1.0 true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 > x2 = [[filter#(_F0, _x2)]] [[times(_x0, plus(_x1, s(_x2)))]] = 0 >= 0 = [[plus(times(_x0, plus(_x1, times(s(_x2), 0))), times(_x0, s(_x2)))]] [[times(_x0, 0)]] = 0 >= 0 = [[0]] [[times(_x0, s(_x1))]] = 0 >= 0 = [[plus(times(_x0, _x1), _x0)]] [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[plus(_x0, s(_x1))]] = x0 >= 0 = [[s(plus(_x0, _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_4, R_0, static, formative) by (P_5, R_0, static, formative), where P_5 consists of: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (P_5, R_0, static, formative) is finite. We consider the dependency pair problem (P_5, R_0, static, 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 : 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, static, formative), (P_2, R_0, static, formative) and (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 with usable rules [Kop12, Thm. 7.44]. (P_3, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: map#(F, cons(X, Y)) >? map#(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 2y1 map# = \G0y1.y1 Using this interpretation, the requirements translate to: [[map#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[map#(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0). 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, static, formative) and (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 with usable rules [Kop12, Thm. 7.44]. (P_2, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: plus#(X, s(Y)) >? plus#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: plus# = \y0y1.y1 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[plus#(_x0, s(_x1))]] = 1 + 2x1 > x1 = [[plus#(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0). 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, static, formative) is finite. We consider the dependency pair problem (P_1, R_0, static, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_1, R_0) are: times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: times#(X, plus(Y, s(Z))) >? times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) >? times#(X, s(Z)) times#(X, s(Y)) >? times#(X, Y) times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 plus = \y0y1.y0 + y1 s = \y0.1 + y0 times = \y0y1.2y0y1 times# = \y0y1.2y1 Using this interpretation, the requirements translate to: [[times#(_x0, plus(_x1, s(_x2)))]] = 2 + 2x1 + 2x2 > 2x1 = [[times#(_x0, plus(_x1, times(s(_x2), 0)))]] [[times#(_x0, plus(_x1, s(_x2)))]] = 2 + 2x1 + 2x2 >= 2 + 2x2 = [[times#(_x0, s(_x2))]] [[times#(_x0, s(_x1))]] = 2 + 2x1 > 2x1 = [[times#(_x0, _x1)]] [[times(_x0, plus(_x1, s(_x2)))]] = 2x0 + 2x0x1 + 2x0x2 >= 2x0 + 2x0x1 + 2x0x2 = [[plus(times(_x0, plus(_x1, times(s(_x2), 0))), times(_x0, s(_x2)))]] [[times(_x0, 0)]] = 0 >= 0 = [[0]] [[times(_x0, s(_x1))]] = 2x0 + 2x0x1 >= x0 + 2x0x1 = [[plus(times(_x0, _x1), _x0)]] [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[plus(_x0, s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 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_6, R_0, static, formative), where P_6 consists of: times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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.