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TRS Standard pair #487073978
details
property
value
status
complete
benchmark
aprove5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Secret_05_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.06013 seconds
cpu usage
5.1046
user time
4.85342
system time
0.251181
max virtual memory
1.8876108E7
max residence set size
428164.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) DependencyGraphProof [EQUIVALENT, 0 ms] (16) AND (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) QReductionProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPOrderProof [EQUIVALENT, 21 ms] (23) QDP (24) PisEmptyProof [EQUIVALENT, 0 ms] (25) YES (26) QDP (27) QDPSizeChangeProof [EQUIVALENT, 0 ms] (28) YES (29) QDP (30) UsableRulesProof [EQUIVALENT, 0 ms] (31) QDP (32) MNOCProof [EQUIVALENT, 0 ms] (33) QDP (34) UsableRulesReductionPairsProof [EQUIVALENT, 15 ms] (35) QDP (36) TransformationProof [EQUIVALENT, 0 ms] (37) QDP (38) UsableRulesProof [EQUIVALENT, 0 ms] (39) QDP (40) QDPOrderProof [EQUIVALENT, 0 ms] (41) QDP (42) PisEmptyProof [EQUIVALENT, 0 ms] (43) YES (44) QDP (45) QDPOrderProof [EQUIVALENT, 15 ms] (46) QDP (47) QDPOrderProof [EQUIVALENT, 0 ms] (48) QDP (49) PisEmptyProof [EQUIVALENT, 0 ms] (50) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, 0) -> x minus(0, y) -> 0 minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) minus(x, plus(y, z)) -> minus(minus(x, y), z) p(s(s(x))) -> s(p(s(x))) p(0) -> s(s(0)) div(s(x), s(y)) -> s(div(minus(x, y), s(y))) div(plus(x, y), z) -> plus(div(x, z), div(y, z)) plus(0, y) -> y plus(s(x), y) -> s(plus(y, minus(s(x), s(0)))) 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: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) MINUS(s(x), s(y)) -> P(s(x)) MINUS(s(x), s(y)) -> P(s(y)) MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z) MINUS(x, plus(y, z)) -> MINUS(x, y) P(s(s(x))) -> P(s(x)) DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) DIV(s(x), s(y)) -> MINUS(x, y) DIV(plus(x, y), z) -> PLUS(div(x, z), div(y, z))
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