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TRS Stand 20472 pair #381711076
details
property
value
status
complete
benchmark
#4.37.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n106.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0949988365173 seconds
cpu usage
0.083641086
max memory
2789376.0
stage attributes
key
value
output-size
5424
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)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: c : [z * z] --> ba f : [ba] --> q g : [ba] --> ca s : [z] --> z 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. The following interpretation satisfies the requirements: c = \y0y1.3y1 g# = \y0.2y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[g#(c(_x0, s(_x1)))]] = 18 + 6x1 > 6x1 = [[g#(c(s(_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_1) by ({}, R_1). 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). This combination (P_1, R_0) has no formative rules! This is exactly 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.
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