details

property	value
status	complete
benchmark	#4.37a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n109.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.117736101151 seconds
cpu usage	0.095286088
max memory	3051520.0

stage attributes

key	value
output-size	5511
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. c : [o * o] --> o f : [o] --> o g : [o] --> o s : [o] --> o f(c(s(X), Y)) => f(c(X, s(Y))) g(c(X, s(Y))) => g(c(s(X), Y)) g(s(f(X))) => g(f(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]): f(c(s(X), Y)) >? f(c(X, s(Y))) g(c(X, s(Y))) >? g(c(s(X), Y)) g(s(f(X))) >? g(f(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = \y0y1.y0 + y1 f = \y0.2 + 2y0 g = \y0.y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[f(c(s(_x0), _x1))]] = 6 + 2x0 + 2x1 >= 6 + 2x0 + 2x1 = [[f(c(_x0, s(_x1)))]] [[g(c(_x0, s(_x1)))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[g(c(s(_x0), _x1))]] [[g(s(f(_x0)))]] = 4 + 2x0 > 2 + 2x0 = [[g(f(_x0))]] We can thus remove the following rules: g(s(f(X))) => g(f(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] f#(c(s(X), Y)) =#> f#(c(X, s(Y))) 1] g#(c(X, s(Y))) =#> g#(c(s(X), Y)) Rules R_0: f(c(s(X), Y)) => f(c(X, s(Y))) g(c(X, s(Y))) => g(c(s(X), Y)) 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 : 0 * 1 : 1 This graph has the following strongly connected components: P_1: f#(c(s(X), Y)) =#> f#(c(X, s(Y))) P_2: g#(c(X, s(Y))) =#> g#(c(s(X), 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) and (P_2, R_0, m, f). 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). This combination (P_2, 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_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, 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: g#(c(X, s(Y))) >? g#(c(s(X), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. popout

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