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TRS Stand 20472 pair #381709961
details
property
value
status
complete
benchmark
IJCAR_18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.98367094994 seconds
cpu usage
4.857890453
max memory
2.88894976E8
stage attributes
key
value
output-size
17211
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) QDPOrderProof [EQUIVALENT, 31 ms] (22) QDP (23) QDPOrderProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) DependencyGraphProof [EQUIVALENT, 0 ms] (28) TRUE (29) QDP (30) QDPSizeChangeProof [EQUIVALENT, 0 ms] (31) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: plus(x, 0) -> x plus(0, y) -> y plus(s(x), y) -> s(plus(x, 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) TIMES(s(x), y) -> PLUS(y, times(x, y)) TIMES(s(x), y) -> TIMES(x, y) DIV(x, y) -> QUOT(x, y, y) QUOT(s(x), s(y), z) -> QUOT(x, y, z) QUOT(x, 0, s(z)) -> DIV(x, s(z)) DIV(div(x, y), z) -> DIV(x, times(y, z)) DIV(div(x, y), z) -> TIMES(y, z) EQ(s(x), s(y)) -> EQ(x, y) DIVIDES(y, x) -> EQ(x, times(div(x, y), y))
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