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TRS Stand 20472 pair #381713867
details
property
value
status
complete
benchmark
jw13.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n081.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.83347392082 seconds
cpu usage
4.27965166
max memory
3.11222272E8
stage attributes
key
value
output-size
2963
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 26 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(f(a, x), f(a, a))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, f(a, x)) -> F(a, f(f(a, x), f(a, a))) F(a, f(a, x)) -> F(f(a, x), f(a, a)) F(a, f(a, x)) -> F(a, a) The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(f(a, x), f(a, a))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, f(a, x)) -> F(a, f(f(a, x), f(a, a))) The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(f(a, x), f(a, a))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. F(a, f(a, x)) -> F(a, f(f(a, x), f(a, a))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO,RATPOLO]: POL(F(x_1, x_2)) = [1/4]x_2 POL(a) = [1/2] POL(f(x_1, x_2)) = [1/2]x_1 The value of delta used in the strict ordering is 1/32. The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: f(a, f(a, x)) -> f(a, f(f(a, x), f(a, a))) ---------------------------------------- (6) Obligation:
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