We consider the system Applicative_AG01_innermost__#4.34. 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, x, y) => x if(false, x, y) => y g(s(x), s(y)) => if(f(x), s(x), s(y)) g(x, c(y)) => g(x, g(s(c(y)), 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#(s(X), s(Y)) =#> if#(f(X), s(X), s(Y)) 2] g#(s(X), s(Y)) =#> f#(X) 3] g#(X, c(Y)) =#> g#(X, g(s(c(Y)), Y)) 4] g#(X, c(Y)) =#> g#(s(c(Y)), Y) 5] map#(F, cons(X, Y)) =#> map#(F, Y) 6] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 7] filter2#(true, F, X, Y) =#> filter#(F, Y) 8] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: f(0) => true f(1) => false f(s(X)) => f(X) if(true, X, Y) => X if(false, X, Y) => Y g(s(X), s(Y)) => if(f(X), s(X), s(Y)) g(X, c(Y)) => g(X, g(s(c(Y)), 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 : 0 * 1 : * 2 : 0 * 3 : 1, 2, 3, 4 * 4 : 1, 2, 3, 4 * 5 : 5 * 6 : 7, 8 * 7 : 6 * 8 : 6 This graph has the following strongly connected components: P_1: f#(s(X)) =#> f#(X) P_2: g#(X, c(Y)) =#> g#(X, g(s(c(Y)), Y)) g#(X, c(Y)) =#> g#(s(c(Y)), 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) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, X, Y) >= X if(false, X, Y) >= Y g(s(X), s(Y)) >= if(f(X), s(X), s(Y)) g(X, c(Y)) >= g(X, g(s(c(Y)), 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 = 3 1 = 3 c = \y0.0 cons = \y0y1.1 + 2y1 f = \y0.3 false = 0 filter = \G0y1.y1 filter2 = \y0G1y2y3.1 + 2y3 filter2# = \y0G1y2y3.1 + 2y3 filter# = \G0y1.2y1 g = \y0y1.0 if = \y0y1y2.y1 + y2 map = \G0y1.y1 nil = 0 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 2 + 4x2 > 1 + 2x2 = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + 2x2 > 2x2 = [[filter#(_F0, _x2)]] [[f(0)]] = 3 >= 0 = [[true]] [[f(1)]] = 3 >= 0 = [[false]] [[f(s(_x0))]] = 3 >= 3 = [[f(_x0)]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[g(s(_x0), s(_x1))]] = 0 >= 0 = [[if(f(_x0), s(_x0), s(_x1))]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[g(_x0, g(s(c(_x1)), _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 a dependency pair problem (P_4, 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), (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]. The usable rules of (P_2, R_0) are: f(0) => true f(1) => false f(s(X)) => f(X) if(true, X, Y) => X if(false, X, Y) => Y g(s(X), s(Y)) => if(f(X), s(X), s(Y)) g(X, c(Y)) => g(X, g(s(c(Y)), Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: g#(X, c(Y)) >? g#(X, g(s(c(Y)), Y)) g#(X, c(Y)) >? g#(s(c(Y)), Y) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, X, Y) >= X if(false, X, Y) >= Y g(s(X), s(Y)) >= if(f(X), s(X), s(Y)) g(X, c(Y)) >= g(X, g(s(c(Y)), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 1 = 3 c = \y0.3 + 3y0 f = \y0.0 false = 0 g = \y0y1.0 g# = \y0y1.3y1 if = \y0y1y2.y1 + y2 s = \y0.0 true = 0 Using this interpretation, the requirements translate to: [[g#(_x0, c(_x1))]] = 9 + 9x1 > 0 = [[g#(_x0, g(s(c(_x1)), _x1))]] [[g#(_x0, c(_x1))]] = 9 + 9x1 > 3x1 = [[g#(s(c(_x1)), _x1)]] [[f(0)]] = 0 >= 0 = [[true]] [[f(1)]] = 0 >= 0 = [[false]] [[f(s(_x0))]] = 0 >= 0 = [[f(_x0)]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[g(s(_x0), s(_x1))]] = 0 >= 0 = [[if(f(_x0), s(_x0), s(_x1))]] [[g(_x0, c(_x1))]] = 0 >= 0 = [[g(_x0, g(s(c(_x1)), _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]. (P_1, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: f#(s(X)) >? f#(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f# = \y0.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[f#(s(_x0))]] = 1 + 2x0 > x0 = [[f#(_x0)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.