We consider the system Applicative_05__Ex5Sorting. Alphabet: 0 : [] --> a ascending!fac6220sort : [b] --> b cons : [a * b] --> b descending!fac6220sort : [b] --> b insert : [a -> a -> a * a -> a -> a * b * a] --> b max : [] --> a -> a -> a min : [] --> a -> a -> a nil : [] --> b s : [a] --> a sort : [a -> a -> a * a -> a -> a * b] --> b Rules: max 0 x => x max x 0 => x max s(x) s(y) => max x y min 0 x => 0 min x 0 => 0 min s(x) s(y) => min x y insert(f, g, nil, x) => cons(x, nil) insert(f, g, cons(x, y), z) => cons(f z x, insert(f, g, y, g z x)) sort(f, g, nil) => nil sort(f, g, cons(x, y)) => insert(f, g, sort(f, g, y), x) ascending!fac6220sort(x) => sort(min, max, x) descending!fac6220sort(x) => sort(max, min, x) 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]). In order to do so, we start by eta-expanding the system, which gives: max(0, X) => X max(X, 0) => X max(s(X), s(Y)) => max(X, Y) min(0, X) => 0 min(X, 0) => 0 min(s(X), s(Y)) => min(X, Y) insert(F, G, nil, X) => cons(X, nil) insert(F, G, cons(X, Y), Z) => cons(F Z X, insert(F, G, Y, G Z X)) sort(F, G, nil) => nil sort(F, G, cons(X, Y)) => insert(F, G, sort(F, G, Y), X) ascending!fac6220sort(X) => sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X) descending!fac6220sort(X) => sort(/\x./\y.max(x, y), /\z./\u.min(z, u), X) We thus obtain the following dependency pair problem (P_0, R_0, static, formative): Dependency Pairs P_0: 0] max#(s(X), s(Y)) =#> max#(X, Y) 1] min#(s(X), s(Y)) =#> min#(X, Y) 2] insert#(F, G, cons(X, Y), Z) =#> insert#(F, G, Y, G Z X) 3] sort#(F, G, cons(X, Y)) =#> insert#(F, G, sort(F, G, Y), X) 4] sort#(F, G, cons(X, Y)) =#> sort#(F, G, Y) 5] ascending!fac6220sort#(X) =#> sort#(/\x./\y.min(x, y), /\z./\u.max(z, u), X) 6] ascending!fac6220sort#(X) =#> min#(Y, Z) 7] ascending!fac6220sort#(X) =#> max#(Y, Z) 8] descending!fac6220sort#(X) =#> sort#(/\x./\y.max(x, y), /\z./\u.min(z, u), X) 9] descending!fac6220sort#(X) =#> max#(Y, Z) 10] descending!fac6220sort#(X) =#> min#(Y, Z) Rules R_0: max(0, X) => X max(X, 0) => X max(s(X), s(Y)) => max(X, Y) min(0, X) => 0 min(X, 0) => 0 min(s(X), s(Y)) => min(X, Y) insert(F, G, nil, X) => cons(X, nil) insert(F, G, cons(X, Y), Z) => cons(F Z X, insert(F, G, Y, G Z X)) sort(F, G, nil) => nil sort(F, G, cons(X, Y)) => insert(F, G, sort(F, G, Y), X) ascending!fac6220sort(X) => sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X) descending!fac6220sort(X) => sort(/\x./\y.max(x, y), /\z./\u.min(z, u), X) 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 : 1 * 2 : 2 * 3 : 2 * 4 : 3, 4 * 5 : 3, 4 * 6 : 1 * 7 : 0 * 8 : 3, 4 * 9 : 0 * 10 : 1 This graph has the following strongly connected components: P_1: max#(s(X), s(Y)) =#> max#(X, Y) P_2: min#(s(X), s(Y)) =#> min#(X, Y) P_3: insert#(F, G, cons(X, Y), Z) =#> insert#(F, G, Y, G Z X) P_4: sort#(F, G, cons(X, Y)) =#> sort#(F, G, 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 with usable rules [Kop12, Thm. 7.44]. (P_4, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: sort#(F, G, cons(X, Y)) >? sort#(F, G, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + 2y1 sort# = \G0G1y2.y2 Using this interpretation, the requirements translate to: [[sort#(_F0, _F1, cons(_x2, _x3))]] = 3 + 2x3 > x3 = [[sort#(_F0, _F1, _x3)]] 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 [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: insert#(F, G, cons(X, Y), Z) >? insert#(F, G, Y, G Z X) max(0, X) >= X max(X, 0) >= X max(s(X), s(Y)) >= max(X, Y) min(0, X) >= 0 min(X, 0) >= 0 min(s(X), s(Y)) >= min(X, Y) insert(F, G, nil, X) >= cons(X, nil) insert(F, G, cons(X, Y), Z) >= cons(F Z X, insert(F, G, Y, G Z X)) sort(F, G, nil) >= nil sort(F, G, cons(X, Y)) >= insert(F, G, sort(F, G, Y), X) ascending!fac6220sort(X) >= sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X) descending!fac6220sort(X) >= sort(/\x./\y.max(x, y), /\z./\u.min(z, u), X) We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( ascending!fac6220sort(X) ) = #argfun-ascending!fac6220sort#(sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X)) pi( descending!fac6220sort(X) ) = #argfun-descending!fac6220sort#(sort(/\x./\y.max(x, y), /\z./\u.min(z, u), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: #argfun-ascending!fac6220sort# = \y0.3 + y0 #argfun-descending!fac6220sort# = \y0.3 + y0 0 = 1 ascending!fac6220sort = \y0.0 cons = \y0y1.1 + y1 descending!fac6220sort = \y0.0 insert = \G0G1y2y3.1 + y2 insert# = \G0G1y2y3.y2 max = \y0y1.y0 + y1 min = \y0y1.3 + y0 nil = 0 s = \y0.y0 sort = \G0G1y2.y2 Using this interpretation, the requirements translate to: [[insert#(_F0, _F1, cons(_x2, _x3), _x4)]] = 1 + x3 > x3 = [[insert#(_F0, _F1, _x3, _F1 _x4 _x2)]] [[max(0, _x0)]] = 1 + x0 >= x0 = [[_x0]] [[max(_x0, 0)]] = 1 + x0 >= x0 = [[_x0]] [[max(s(_x0), s(_x1))]] = x0 + x1 >= x0 + x1 = [[max(_x0, _x1)]] [[min(0, _x0)]] = 4 >= 1 = [[0]] [[min(_x0, 0)]] = 3 + x0 >= 1 = [[0]] [[min(s(_x0), s(_x1))]] = 3 + x0 >= 3 + x0 = [[min(_x0, _x1)]] [[insert(_F0, _F1, nil, _x2)]] = 1 >= 1 = [[cons(_x2, nil)]] [[insert(_F0, _F1, cons(_x2, _x3), _x4)]] = 2 + x3 >= 2 + x3 = [[cons(_F0 _x4 _x2, insert(_F0, _F1, _x3, _F1 _x4 _x2))]] [[sort(_F0, _F1, nil)]] = 0 >= 0 = [[nil]] [[sort(_F0, _F1, cons(_x2, _x3))]] = 1 + x3 >= 1 + x3 = [[insert(_F0, _F1, sort(_F0, _F1, _x3), _x2)]] [[#argfun-ascending!fac6220sort#(sort(/\x./\y.min(x, y), /\z./\u.max(z, u), _x0))]] = 3 + x0 >= x0 = [[sort(/\x./\y.min(x, y), /\z./\u.max(z, u), _x0)]] [[#argfun-descending!fac6220sort#(sort(/\x./\y.max(x, y), /\z./\u.min(z, u), _x0))]] = 3 + x0 >= x0 = [[sort(/\x./\y.max(x, y), /\z./\u.min(z, u), _x0)]] 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: min#(s(X), s(Y)) >? min#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: min# = \y0y1.y0 + y1 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[min#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[min#(_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]. (P_1, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: max#(s(X), s(Y)) >? max#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: max# = \y0y1.y0 + y1 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[max#(s(_x0), s(_x1))]] = 6 + 2x0 + 2x1 > x0 + x1 = [[max#(_x0, _x1)]] 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.