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SRS Stand 10685 pair #381718084
details
property
value
status
complete
benchmark
aprove08.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n002.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.47996807098 seconds
cpu usage
17.543721592
max memory
1.974542336E9
stage attributes
key
value
output-size
9300
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 64 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 11 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 2 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 59 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(0(x1)) -> p(s(p(s(0(p(s(p(s(x1))))))))) i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(0(x1)) -> p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(i(x_1)) = 1 + x_1 POL(j(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: i(0(x1)) -> p(s(p(s(0(p(s(p(s(x1))))))))) j(0(x1)) -> p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: i(s(x1)) -> p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) j(s(x1)) -> s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(x1))))))))) Q is empty. ---------------------------------------- (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:
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