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TRS Standard pair #487068798
details
property
value
status
complete
benchmark
IJCAR_26a.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.28501 seconds
cpu usage
5.61654
user time
5.36372
system time
0.252813
max virtual memory
1.874378E7
max residence set size
370512.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 22 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) MRRProof [EQUIVALENT, 9 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) TRUE (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPOrderProof [EQUIVALENT, 42 ms] (26) QDP (27) QDPOrderProof [EQUIVALENT, 11 ms] (28) QDP (29) TransformationProof [EQUIVALENT, 0 ms] (30) QDP (31) DependencyGraphProof [EQUIVALENT, 0 ms] (32) TRUE (33) QDP (34) QDPSizeChangeProof [EQUIVALENT, 0 ms] (35) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0) -> 0 p(s(x)) -> x plus(x, 0) -> x plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) plus(s(x), y) -> s(plus(p(s(x)), y)) plus(x, s(y)) -> s(plus(x, p(s(y)))) times(0, y) -> 0 times(s(0), y) -> y times(s(x), y) -> plus(y, times(x, y)) div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) div(div(x, y), z) -> div(x, times(y, z)) eq(0, 0) -> true eq(s(x), 0) -> false eq(0, s(y)) -> false eq(s(x), s(y)) -> eq(x, y) divides(y, x) -> eq(x, times(div(x, y), y)) prime(s(s(x))) -> pr(s(s(x)), s(x)) pr(x, s(0)) -> true pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y)) if(true, x, y) -> false if(false, x, y) -> pr(x, y) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), y) -> PLUS(x, y) PLUS(s(x), y) -> PLUS(p(s(x)), y) PLUS(s(x), y) -> P(s(x)) PLUS(x, s(y)) -> PLUS(x, p(s(y))) PLUS(x, s(y)) -> P(s(y)) TIMES(s(x), y) -> PLUS(y, times(x, y)) TIMES(s(x), y) -> TIMES(x, y) DIV(x, y) -> QUOT(x, y, y)
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