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TRS Stand 20472 pair #381710482
details
property
value
status
complete
benchmark
#4.17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n055.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.13007307053 seconds
cpu usage
4.756814665
max memory
3.19619072E8
stage attributes
key
value
output-size
5270
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) NonTerminationLoopProof [COMPLETE, 0 ms] (14) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(g(x), s(0), y) -> f(g(s(0)), y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 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(g(x), s(0), y) -> F(g(s(0)), y, g(x)) F(g(x), s(0), y) -> G(s(0)) G(s(x)) -> G(x) The TRS R consists of the following rules: f(g(x), s(0), y) -> f(g(s(0)), y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 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 2 SCCs with 1 less node. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: G(s(x)) -> G(x) The TRS R consists of the following rules: f(g(x), s(0), y) -> f(g(s(0)), y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 0 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation:
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