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TRS Standard pair #487069163
details
property
value
status
complete
benchmark
cade06.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n177.star.cs.uiowa.edu
space
GTSSK07
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.58394 seconds
cpu usage
6.20524
user time
5.9285
system time
0.276738
max virtual memory
1.8477524E7
max residence set size
372592.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 25 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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 0 ms] (25) QDP (26) NonInfProof [EQUIVALENT, 120 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(true, x, y, z) -> f(gt(x, plus(y, z)), x, s(y), z) f(true, x, y, z) -> f(gt(x, plus(y, z)), x, y, s(z)) plus(n, 0) -> n plus(n, s(m)) -> s(plus(n, m)) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is plus(n, 0) -> n plus(n, s(m)) -> s(plus(n, m)) gt(0, v) -> false gt(s(u), 0) -> true gt(s(u), s(v)) -> gt(u, v) The TRS R 2 is f(true, x, y, z) -> f(gt(x, plus(y, z)), x, s(y), z) f(true, x, y, z) -> f(gt(x, plus(y, z)), x, y, s(z)) The signature Sigma is {f_4} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(true, x, y, z) -> f(gt(x, plus(y, z)), x, s(y), z) f(true, x, y, z) -> f(gt(x, plus(y, z)), x, y, s(z)) plus(n, 0) -> n plus(n, s(m)) -> s(plus(n, m)) 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: f(true, x0, x1, x2) plus(x0, 0) plus(x0, s(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.
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