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TRS Standard pair #516965275
details
property
value
status
complete
benchmark
enger-nonloop-unbounded.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n098.star.cs.uiowa.edu
space
EEG_IJCAR_12
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
12.4931139946 seconds
cpu usage
40.685690061
max memory
1.37213952E9
stage attributes
key
value
output-size
8282
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 1 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonLoopProof [COMPLETE, 44 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(true, x, y) -> h(gt(x, y), s(x), s(y)) gt(0, x) -> false gt(s(x), 0) -> true gt(s(x), s(y)) -> gt(x, y) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is gt(0, x) -> false gt(s(x), 0) -> true gt(s(x), s(y)) -> gt(x, y) The TRS R 2 is h(true, x, y) -> h(gt(x, y), s(x), s(y)) The signature Sigma is {h_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(true, x, y) -> h(gt(x, y), s(x), s(y)) gt(0, x) -> false gt(s(x), 0) -> true gt(s(x), s(y)) -> gt(x, y) The set Q consists of the following terms: h(true, x0, x1) gt(0, x0) gt(s(x0), 0) gt(s(x0), s(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: H(true, x, y) -> H(gt(x, y), s(x), s(y)) H(true, x, y) -> GT(x, y) GT(s(x), s(y)) -> GT(x, y) The TRS R consists of the following rules: h(true, x, y) -> h(gt(x, y), s(x), s(y)) gt(0, x) -> false gt(s(x), 0) -> true
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