/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. active : [o] --> o c : [] --> o d : [] --> o g : [o] --> o h : [o] --> o mark : [o] --> o active(g(X)) => mark(h(X)) active(c) => mark(d) active(h(d)) => mark(g(c)) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(X)) mark(c) => active(c) mark(d) => active(d) g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(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(g(X)) >? mark(h(X)) active(c) >? mark(d) active(h(d)) >? mark(g(c)) mark(g(X)) >? active(g(X)) mark(h(X)) >? active(h(X)) mark(c) >? active(c) mark(d) >? active(d) g(mark(X)) >? g(X) g(active(X)) >? g(X) h(mark(X)) >? h(X) h(active(X)) >? h(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active = \y0.1 + y0 c = 0 d = 0 g = \y0.2y0 h = \y0.y0 mark = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[active(g(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(h(_x0))]] [[active(c)]] = 1 >= 1 = [[mark(d)]] [[active(h(d))]] = 1 >= 1 = [[mark(g(c))]] [[mark(g(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[active(g(_x0))]] [[mark(h(_x0))]] = 1 + 2x0 >= 1 + x0 = [[active(h(_x0))]] [[mark(c)]] = 1 >= 1 = [[active(c)]] [[mark(d)]] = 1 >= 1 = [[active(d)]] [[g(mark(_x0))]] = 2 + 4x0 > 2x0 = [[g(_x0)]] [[g(active(_x0))]] = 2 + 2x0 > 2x0 = [[g(_x0)]] [[h(mark(_x0))]] = 1 + 2x0 > x0 = [[h(_x0)]] [[h(active(_x0))]] = 1 + x0 > x0 = [[h(_x0)]] We can thus remove the following rules: g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(X) 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#(g(X)) =#> mark#(h(X)) 1] active#(c) =#> mark#(d) 2] active#(h(d)) =#> mark#(g(c)) 3] mark#(g(X)) =#> active#(g(X)) 4] mark#(h(X)) =#> active#(h(X)) 5] mark#(c) =#> active#(c) 6] mark#(d) =#> active#(d) Rules R_0: active(g(X)) => mark(h(X)) active(c) => mark(d) active(h(d)) => mark(g(c)) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(X)) mark(c) => active(c) mark(d) => active(d) 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 : 4 * 1 : 6 * 2 : 3 * 3 : 0 * 4 : 2 * 5 : 1 * 6 : This graph has the following strongly connected components: P_1: active#(g(X)) =#> mark#(h(X)) active#(h(d)) =#> mark#(g(c)) mark#(g(X)) =#> active#(g(X)) mark#(h(X)) =#> active#(h(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). 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). This combination (P_1, R_0) has no formative rules! We will name the empty set of rules:R_1. By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, 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: active#(g(X)) >? mark#(h(X)) active#(h(d)) >? mark#(g(c)) mark#(g(X)) >? active#(g(X)) mark#(h(X)) >? active#(h(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active# = \y0.y0 c = 0 d = 3 g = \y0.2y0 h = \y0.2y0 mark# = \y0.y0 Using this interpretation, the requirements translate to: [[active#(g(_x0))]] = 2x0 >= 2x0 = [[mark#(h(_x0))]] [[active#(h(d))]] = 6 > 0 = [[mark#(g(c))]] [[mark#(g(_x0))]] = 2x0 >= 2x0 = [[active#(g(_x0))]] [[mark#(h(_x0))]] = 2x0 >= 2x0 = [[active#(h(_x0))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of: active#(g(X)) =#> mark#(h(X)) mark#(g(X)) =#> active#(g(X)) mark#(h(X)) =#> active#(h(X)) Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, 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 : 2 * 1 : 0 * 2 : 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. [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.