details

property	value
status	complete
benchmark	tricky1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n108.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0879790782928 seconds
cpu usage	0.075369483
max memory	3633152.0

stage attributes

key	value
output-size	3972
starexec-result	YES

output

/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 f : [o * o] --> o g : [o] --> o p : [o] --> o s : [o] --> o f(g(X), g(Y)) => f(p(f(g(X), s(Y))), g(s(p(X)))) p(0) => g(0) g(s(p(X))) => p(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#(g(X), g(Y)) =#> f#(p(f(g(X), s(Y))), g(s(p(X)))) 1] f#(g(X), g(Y)) =#> p#(f(g(X), s(Y))) 2] f#(g(X), g(Y)) =#> f#(g(X), s(Y)) 3] f#(g(X), g(Y)) =#> g#(X) 4] f#(g(X), g(Y)) =#> g#(s(p(X))) 5] f#(g(X), g(Y)) =#> p#(X) 6] p#(0) =#> g#(0) 7] g#(s(p(X))) =#> p#(X) Rules R_0: f(g(X), g(Y)) => f(p(f(g(X), s(Y))), g(s(p(X)))) p(0) => g(0) g(s(p(X))) => p(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 : 0, 1, 2, 3, 4, 5 * 1 : * 2 : * 3 : 7 * 4 : 7 * 5 : 6 * 6 : * 7 : 6 This graph has the following strongly connected components: P_1: f#(g(X), g(Y)) =#> f#(p(f(g(X), s(Y))), g(s(p(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). The formative rules of (P_1, R_0) are R_1 ::= p(0) => g(0) g(s(p(X))) => p(X) 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: f#(g(X), g(Y)) >? f#(p(f(g(X), s(Y))), g(s(p(X)))) p(0) >= g(0) g(s(p(X))) >= p(X) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: f(x_1,x_2) = f(x_2) f#(x_1,x_2) = f#(x_1) This leaves the following ordering requirements: f#(g(X), g(Y)) > f#(p(f(g(X), s(Y))), g(s(p(X)))) The following interpretation satisfies the requirements: 0 = 0 popout

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