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TRS Stand 20472 pair #381713562
details
property
value
status
complete
benchmark
#4.7.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n096.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.96620798111 seconds
cpu usage
4.160634726
max memory
2.99687936E8
stage attributes
key
value
output-size
3826
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 3 ms] (8) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x)) -> f(g(x, x)) g(0, 1) -> s(0) 0 -> 1 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(s(x)) -> F(g(x, x)) F(s(x)) -> G(x, x) The TRS R consists of the following rules: f(s(x)) -> f(g(x, x)) g(0, 1) -> s(0) 0 -> 1 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(g(x, x)) The TRS R consists of the following rules: f(s(x)) -> f(g(x, x)) g(0, 1) -> s(0) 0 -> 1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. No dependency pairs are removed. The following rules are removed from R: f(s(x)) -> f(g(x, x)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(1) = 0 POL(F(x_1)) = 2*x_1 POL(g(x_1, x_2)) = x_1 + x_2 POL(s(x_1)) = 2*x_1 ----------------------------------------
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