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SRS_Standard 2019-03-29 03.29 pair #432294025
details
property
value
status
complete
benchmark
aprove09.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
Secret_06_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
8.34583 seconds
cpu usage
56.7642
user time
54.3056
system time
2.45867
max virtual memory
2.1012172E7
max residence set size
3727160.0
stage attributes
key
value
starexec-result
YES
output
28.29/8.24 YES 28.29/8.25 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 28.29/8.25 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 28.29/8.25 28.29/8.25 28.29/8.25 Termination w.r.t. Q of the given QTRS could be proven: 28.29/8.25 28.29/8.25 (0) QTRS 28.29/8.25 (1) QTRSRRRProof [EQUIVALENT, 68 ms] 28.29/8.25 (2) QTRS 28.29/8.25 (3) DependencyPairsProof [EQUIVALENT, 35 ms] 28.29/8.25 (4) QDP 28.29/8.25 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 28.29/8.25 (6) AND 28.29/8.25 (7) QDP 28.29/8.25 (8) UsableRulesProof [EQUIVALENT, 0 ms] 28.29/8.25 (9) QDP 28.29/8.25 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 28.29/8.25 (11) YES 28.29/8.25 (12) QDP 28.29/8.25 (13) MNOCProof [EQUIVALENT, 0 ms] 28.29/8.25 (14) QDP 28.29/8.25 (15) UsableRulesProof [EQUIVALENT, 0 ms] 28.29/8.25 (16) QDP 28.29/8.25 (17) QReductionProof [EQUIVALENT, 0 ms] 28.29/8.25 (18) QDP 28.29/8.25 (19) QDPOrderProof [EQUIVALENT, 34 ms] 28.29/8.25 (20) QDP 28.29/8.25 (21) PisEmptyProof [EQUIVALENT, 0 ms] 28.29/8.25 (22) YES 28.29/8.25 28.29/8.25 28.29/8.25 ---------------------------------------- 28.29/8.25 28.29/8.25 (0) 28.29/8.25 Obligation: 28.29/8.25 Q restricted rewrite system: 28.29/8.25 The TRS R consists of the following rules: 28.29/8.25 28.29/8.25 q(0(x1)) -> p(p(s(s(0(s(s(s(s(x1))))))))) 28.29/8.25 q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))) 28.29/8.25 r(0(x1)) -> p(s(p(s(0(p(p(p(s(s(s(x1))))))))))) 28.29/8.25 r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1)))))))))) 28.29/8.25 p(p(s(x1))) -> p(x1) 28.29/8.25 p(s(x1)) -> x1 28.29/8.25 p(0(x1)) -> 0(s(s(s(x1)))) 28.29/8.25 28.29/8.25 Q is empty. 28.29/8.25 28.29/8.25 ---------------------------------------- 28.29/8.25 28.29/8.25 (1) QTRSRRRProof (EQUIVALENT) 28.29/8.25 Used ordering: 28.29/8.25 Polynomial interpretation [POLO]: 28.29/8.25 28.29/8.25 POL(0(x_1)) = x_1 28.29/8.25 POL(p(x_1)) = x_1 28.29/8.25 POL(q(x_1)) = 1 + x_1 28.29/8.25 POL(r(x_1)) = 1 + x_1 28.29/8.25 POL(s(x_1)) = x_1 28.29/8.25 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 28.29/8.25 28.29/8.25 q(0(x1)) -> p(p(s(s(0(s(s(s(s(x1))))))))) 28.29/8.25 r(0(x1)) -> p(s(p(s(0(p(p(p(s(s(s(x1))))))))))) 28.29/8.25 28.29/8.25 28.29/8.25 28.29/8.25 28.29/8.25 ---------------------------------------- 28.29/8.25 28.29/8.25 (2) 28.29/8.25 Obligation: 28.29/8.25 Q restricted rewrite system: 28.29/8.25 The TRS R consists of the following rules: 28.29/8.25 28.29/8.25 q(s(x1)) -> p(p(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))) 28.29/8.25 r(s(x1)) -> p(s(p(s(s(q(p(s(p(s(x1)))))))))) 28.29/8.25 p(p(s(x1))) -> p(x1) 28.29/8.25 p(s(x1)) -> x1 28.29/8.25 p(0(x1)) -> 0(s(s(s(x1)))) 28.29/8.25 28.29/8.25 Q is empty. 28.29/8.25 28.29/8.25 ---------------------------------------- 28.29/8.25 28.29/8.25 (3) DependencyPairsProof (EQUIVALENT) 28.29/8.25 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 28.29/8.25 ---------------------------------------- 28.29/8.25 28.29/8.25 (4) 28.29/8.25 Obligation: 28.29/8.25 Q DP problem: 28.29/8.25 The TRS P consists of the following rules: 28.29/8.25 28.29/8.25 Q(s(x1)) -> P(p(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))) 28.29/8.25 Q(s(x1)) -> P(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))) 28.29/8.25 Q(s(x1)) -> R(p(p(s(s(x1))))) 28.29/8.25 Q(s(x1)) -> P(p(s(s(x1)))) 28.29/8.25 Q(s(x1)) -> P(s(s(x1))) 28.29/8.25 R(s(x1)) -> P(s(p(s(s(q(p(s(p(s(x1))))))))))
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