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TRS Stand 20472 pair #381709622
details
property
value
status
complete
benchmark
cade01.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
GTSSK07
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.23976898193 seconds
cpu usage
5.019773796
max memory
3.28192E8
stage attributes
key
value
output-size
11408
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) NonInfProof [EQUIVALENT, 0 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, y) -> cond(gt(x, y), x, y) cond(false, x, y) -> 0 cond(true, x, y) -> s(minus(x, s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, y) -> cond(gt(x, y), x, y) cond(false, x, y) -> 0 cond(true, x, y) -> s(minus(x, s(y))) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The set Q consists of the following terms: minus(x0, x1) cond(false, x0, x1) cond(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: MINUS(x, y) -> COND(gt(x, y), x, y) MINUS(x, y) -> GT(x, y) COND(true, x, y) -> MINUS(x, s(y)) GT(s(u), s(v)) -> GT(u, v) The TRS R consists of the following rules:
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