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TRS Stand 20472 pair #381712787
details
property
value
status
complete
benchmark
test4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n077.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.69088482857 seconds
cpu usage
3.862558242
max memory
2.39915008E8
stage attributes
key
value
output-size
5156
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 6 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, a) -> f(a, b) f(a, b) -> f(s(a), c) f(s(X), c) -> f(X, c) f(c, c) -> f(a, a) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is f(a, a) -> f(a, b) f(a, b) -> f(s(a), c) f(s(X), c) -> f(X, c) f(c, c) -> f(a, a) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, a) -> f(a, b) f(a, b) -> f(s(a), c) f(s(X), c) -> f(X, c) f(c, c) -> f(a, a) The set Q consists of the following terms: f(a, a) f(a, b) f(s(x0), c) f(c, c) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, a) -> F(a, b) F(a, b) -> F(s(a), c) F(s(X), c) -> F(X, c) F(c, c) -> F(a, a) The TRS R consists of the following rules: f(a, a) -> f(a, b) f(a, b) -> f(s(a), c) f(s(X), c) -> f(X, c) f(c, c) -> f(a, a) The set Q consists of the following terms:
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