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TRS Standard pair #516964010
details
property
value
status
complete
benchmark
11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n171.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.3708729744 seconds
cpu usage
6.359923587
max memory
4.88542208E8
stage attributes
key
value
output-size
6883
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms]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) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 3 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) TransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is h(0) -> 0 h(g(x, y)) -> y The TRS R 2 is f(0, 1, x) -> f(h(x), h(x), x) The signature Sigma is {f_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y The set Q consists of the following terms: f(0, 1, x0) h(0) h(g(x0, x1)) ---------------------------------------- (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(0, 1, x) -> F(h(x), h(x), x) F(0, 1, x) -> H(x) The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y The set Q consists of the following terms:
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