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TRS Standard pair #516967680
details
property
value
status
complete
benchmark
#4.26.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.50914216042 seconds
cpu usage
6.532016367
max memory
4.86334464E8
stage attributes
key
value
output-size
9293
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 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) QDPOrderProof [EQUIVALENT, 35 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0) -> 0 p(s(x)) -> x le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) minus(x, y) -> if(le(x, y), x, y) if(true, x, y) -> 0 if(false, x, y) -> s(minus(p(x), y)) 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: p(0) -> 0 p(s(x)) -> x le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) minus(x, y) -> if(le(x, y), x, y) if(true, x, y) -> 0 if(false, x, y) -> s(minus(p(x), y)) The set Q consists of the following terms: p(0) p(s(x0)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) minus(x0, x1) if(true, x0, x1) if(false, 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: LE(s(x), s(y)) -> LE(x, y)
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