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TRS Stand 20472 pair #381716002
details
property
value
status
complete
benchmark
LISTUTILITIES_nosorts_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n040.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.89829397202 seconds
cpu usage
4.183444248
max memory
2.07798272E8
stage attributes
key
value
output-size
14139
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) afterNth(N, XS) -> snd(splitAt(N, XS)) and(tt, X) -> activate(X) fst(pair(X, Y)) -> X head(cons(N, XS)) -> N natsFrom(N) -> cons(N, n__natsFrom(s(N))) sel(N, XS) -> head(afterNth(N, XS)) snd(pair(X, Y)) -> Y splitAt(0, XS) -> pair(nil, XS) splitAt(s(N), cons(X, XS)) -> U11(tt, N, X, activate(XS)) tail(cons(N, XS)) -> activate(XS) take(N, XS) -> fst(splitAt(N, XS)) natsFrom(X) -> n__natsFrom(X) activate(n__natsFrom(X)) -> natsFrom(X) activate(X) -> X 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: U11^1(tt, N, X, XS) -> U12^1(splitAt(activate(N), activate(XS)), activate(X)) U11^1(tt, N, X, XS) -> SPLITAT(activate(N), activate(XS)) U11^1(tt, N, X, XS) -> ACTIVATE(N) U11^1(tt, N, X, XS) -> ACTIVATE(XS) U11^1(tt, N, X, XS) -> ACTIVATE(X) U12^1(pair(YS, ZS), X) -> ACTIVATE(X) AFTERNTH(N, XS) -> SND(splitAt(N, XS)) AFTERNTH(N, XS) -> SPLITAT(N, XS) AND(tt, X) -> ACTIVATE(X) SEL(N, XS) -> HEAD(afterNth(N, XS)) SEL(N, XS) -> AFTERNTH(N, XS) SPLITAT(s(N), cons(X, XS)) -> U11^1(tt, N, X, activate(XS)) SPLITAT(s(N), cons(X, XS)) -> ACTIVATE(XS) TAIL(cons(N, XS)) -> ACTIVATE(XS) TAKE(N, XS) -> FST(splitAt(N, XS)) TAKE(N, XS) -> SPLITAT(N, XS) ACTIVATE(n__natsFrom(X)) -> NATSFROM(X) The TRS R consists of the following rules: U11(tt, N, X, XS) -> U12(splitAt(activate(N), activate(XS)), activate(X)) U12(pair(YS, ZS), X) -> pair(cons(activate(X), YS), ZS) afterNth(N, XS) -> snd(splitAt(N, XS)) and(tt, X) -> activate(X) fst(pair(X, Y)) -> X head(cons(N, XS)) -> N natsFrom(N) -> cons(N, n__natsFrom(s(N))) sel(N, XS) -> head(afterNth(N, XS)) snd(pair(X, Y)) -> Y
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