/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o active : [o] --> o cons : [o * o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o tail : [o] --> o top : [o] --> o zeros : [] --> o active(zeros) => mark(cons(0, zeros)) active(tail(cons(X, Y))) => mark(Y) active(cons(X, Y)) => cons(active(X), Y) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(zeros) => ok(zeros) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(0) => ok(0) proper(tail(X)) => tail(proper(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) tail(ok(X)) => ok(tail(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(X)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): active(zeros) >? mark(cons(0, zeros)) active(tail(cons(X, Y))) >? mark(Y) active(cons(X, Y)) >? cons(active(X), Y) active(tail(X)) >? tail(active(X)) cons(mark(X), Y) >? mark(cons(X, Y)) tail(mark(X)) >? mark(tail(X)) proper(zeros) >? ok(zeros) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(0) >? ok(0) proper(tail(X)) >? tail(proper(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) tail(ok(X)) >? ok(tail(X)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 active = \y0.y0 cons = \y0y1.y1 + 2y0 mark = \y0.y0 ok = \y0.y0 proper = \y0.y0 tail = \y0.2 + y0 top = \y0.y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(tail(cons(_x0, _x1)))]] = 2 + x1 + 2x0 > x1 = [[mark(_x1)]] [[active(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(active(_x0), _x1)]] [[active(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(active(_x0))]] [[cons(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[tail(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(tail(_x0))]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(proper(_x0))]] [[cons(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(cons(_x0, _x1))]] [[tail(ok(_x0))]] = 2 + x0 >= 2 + x0 = [[ok(tail(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: active(tail(cons(X, Y))) => mark(Y) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] active#(zeros) =#> cons#(0, zeros) 1] active#(cons(X, Y)) =#> cons#(active(X), Y) 2] active#(cons(X, Y)) =#> active#(X) 3] active#(tail(X)) =#> tail#(active(X)) 4] active#(tail(X)) =#> active#(X) 5] cons#(mark(X), Y) =#> cons#(X, Y) 6] tail#(mark(X)) =#> tail#(X) 7] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y)) 8] proper#(cons(X, Y)) =#> proper#(X) 9] proper#(cons(X, Y)) =#> proper#(Y) 10] proper#(tail(X)) =#> tail#(proper(X)) 11] proper#(tail(X)) =#> proper#(X) 12] cons#(ok(X), ok(Y)) =#> cons#(X, Y) 13] tail#(ok(X)) =#> tail#(X) 14] top#(mark(X)) =#> top#(proper(X)) 15] top#(mark(X)) =#> proper#(X) 16] top#(ok(X)) =#> top#(active(X)) 17] top#(ok(X)) =#> active#(X) Rules R_0: active(zeros) => mark(cons(0, zeros)) active(cons(X, Y)) => cons(active(X), Y) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(zeros) => ok(zeros) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(0) => ok(0) proper(tail(X)) => tail(proper(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) tail(ok(X)) => ok(tail(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(X)) 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 : * 1 : 5, 12 * 2 : 0, 1, 2, 3, 4 * 3 : 6, 13 * 4 : 0, 1, 2, 3, 4 * 5 : 5, 12 * 6 : 6, 13 * 7 : 5, 12 * 8 : 7, 8, 9, 10, 11 * 9 : 7, 8, 9, 10, 11 * 10 : 6, 13 * 11 : 7, 8, 9, 10, 11 * 12 : 5, 12 * 13 : 6, 13 * 14 : 14, 15, 16, 17 * 15 : 7, 8, 9, 10, 11 * 16 : 14, 15, 16, 17 * 17 : 0, 1, 2, 3, 4 This graph has the following strongly connected components: P_1: active#(cons(X, Y)) =#> active#(X) active#(tail(X)) =#> active#(X) P_2: cons#(mark(X), Y) =#> cons#(X, Y) cons#(ok(X), ok(Y)) =#> cons#(X, Y) P_3: tail#(mark(X)) =#> tail#(X) tail#(ok(X)) =#> tail#(X) P_4: proper#(cons(X, Y)) =#> proper#(X) proper#(cons(X, Y)) =#> proper#(Y) proper#(tail(X)) =#> proper#(X) P_5: top#(mark(X)) =#> top#(proper(X)) top#(ok(X)) =#> top#(active(X)) 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), (P_4, R_0, m, f) and (P_5, 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), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). The formative rules of (P_5, R_0) are R_1 ::= active(zeros) => mark(cons(0, zeros)) active(cons(X, Y)) => cons(active(X), Y) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(zeros) => ok(zeros) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(0) => ok(0) proper(tail(X)) => tail(proper(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) tail(ok(X)) => ok(tail(X)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_5, R_0, minimal, formative) by (P_5, 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), (P_4, 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]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: top#(mark(X)) >? top#(proper(X)) top#(ok(X)) >? top#(active(X)) active(zeros) >= mark(cons(0, zeros)) active(cons(X, Y)) >= cons(active(X), Y) active(tail(X)) >= tail(active(X)) cons(mark(X), Y) >= mark(cons(X, Y)) tail(mark(X)) >= mark(tail(X)) proper(zeros) >= ok(zeros) proper(cons(X, Y)) >= cons(proper(X), proper(Y)) proper(0) >= ok(0) proper(tail(X)) >= tail(proper(X)) cons(ok(X), ok(Y)) >= ok(cons(X, Y)) tail(ok(X)) >= ok(tail(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 active = \y0.2y0 cons = \y0y1.y0 mark = \y0.1 + 2y0 ok = \y0.1 + 2y0 proper = \y0.1 + 2y0 tail = \y0.y0 top# = \y0.y0 zeros = 2 Using this interpretation, the requirements translate to: [[top#(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[top#(proper(_x0))]] [[top#(ok(_x0))]] = 1 + 2x0 > 2x0 = [[top#(active(_x0))]] [[active(zeros)]] = 4 >= 1 = [[mark(cons(0, zeros))]] [[active(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(active(_x0), _x1)]] [[active(tail(_x0))]] = 2x0 >= 2x0 = [[tail(active(_x0))]] [[cons(mark(_x0), _x1)]] = 1 + 2x0 >= 1 + 2x0 = [[mark(cons(_x0, _x1))]] [[tail(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(tail(_x0))]] [[proper(zeros)]] = 5 >= 5 = [[ok(zeros)]] [[proper(cons(_x0, _x1))]] = 1 + 2x0 >= 1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(0)]] = 1 >= 1 = [[ok(0)]] [[proper(tail(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[tail(proper(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(cons(_x0, _x1))]] [[tail(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(tail(_x0))]] 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: top#(mark(X)) =#> top#(proper(X)) 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), (P_4, 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). The formative rules of (P_6, R_1) are R_2 ::= active(zeros) => mark(cons(0, zeros)) active(cons(X, Y)) => cons(active(X), Y) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(tail(X)) => tail(proper(X)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_6, R_1, minimal, formative) by (P_6, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_6, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_2, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_6, R_2) are: cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(tail(X)) => tail(proper(X)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: top#(mark(X)) >? top#(proper(X)) cons(mark(X), Y) >= mark(cons(X, Y)) tail(mark(X)) >= mark(tail(X)) proper(cons(X, Y)) >= cons(proper(X), proper(Y)) proper(tail(X)) >= tail(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.y0 mark = \y0.3 + 3y0 proper = \y0.0 tail = \y0.3y0 top# = \y0.3y0 Using this interpretation, the requirements translate to: [[top#(mark(_x0))]] = 9 + 9x0 > 0 = [[top#(proper(_x0))]] [[cons(mark(_x0), _x1)]] = 3 + 3x0 >= 3 + 3x0 = [[mark(cons(_x0, _x1))]] [[tail(mark(_x0))]] = 9 + 9x0 >= 3 + 9x0 = [[mark(tail(_x0))]] [[proper(cons(_x0, _x1))]] = 0 >= 0 = [[cons(proper(_x0), proper(_x1))]] [[proper(tail(_x0))]] = 0 >= 0 = [[tail(proper(_x0))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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), (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). We apply the subterm criterion with the following projection function: nu(proper#) = 1 Thus, we can orient the dependency pairs as follows: nu(proper#(cons(X, Y))) = cons(X, Y) |> X = nu(proper#(X)) nu(proper#(cons(X, Y))) = cons(X, Y) |> Y = nu(proper#(Y)) nu(proper#(tail(X))) = tail(X) |> X = nu(proper#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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 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(tail#) = 1 Thus, we can orient the dependency pairs as follows: nu(tail#(mark(X))) = mark(X) |> X = nu(tail#(X)) nu(tail#(ok(X))) = ok(X) |> X = nu(tail#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, 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 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(cons#) = 1 Thus, we can orient the dependency pairs as follows: nu(cons#(mark(X), Y)) = mark(X) |> X = nu(cons#(X, Y)) nu(cons#(ok(X), ok(Y))) = ok(X) |> X = nu(cons#(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(active#) = 1 Thus, we can orient the dependency pairs as follows: nu(active#(cons(X, Y))) = cons(X, Y) |> X = nu(active#(X)) nu(active#(tail(X))) = tail(X) |> X = nu(active#(X)) 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. [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.