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TRS Stand 20472 pair #381712680
details
property
value
status
complete
benchmark
26.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.80371403694 seconds
cpu usage
4.141773433
max memory
2.04017664E8
stage attributes
key
value
output-size
3237
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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, 0 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(f(x)) -> f(g(f(x), x)) f(f(x)) -> f(h(f(x), f(x))) g(x, y) -> y h(x, x) -> g(x, 0) 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(f(x)) -> F(g(f(x), x)) F(f(x)) -> G(f(x), x) F(f(x)) -> F(h(f(x), f(x))) F(f(x)) -> H(f(x), f(x)) H(x, x) -> G(x, 0) The TRS R consists of the following rules: f(f(x)) -> f(g(f(x), x)) f(f(x)) -> f(h(f(x), f(x))) g(x, y) -> y h(x, x) -> g(x, 0) 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 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x)) -> F(h(f(x), f(x))) F(f(x)) -> F(g(f(x), x)) The TRS R consists of the following rules: f(f(x)) -> f(g(f(x), x)) f(f(x)) -> f(h(f(x), f(x))) g(x, y) -> y h(x, x) -> g(x, 0) 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(f(x)) -> F(h(f(x), f(x))) F(f(x)) -> F(g(f(x), x)) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. F(x1) = x1
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