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TRS Stand 20472 pair #381716157
details
property
value
status
complete
benchmark
Ex3_3_25_Bor03_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.27815294266 seconds
cpu usage
6.041817479
max memory
3.9831552E8
stage attributes
key
value
output-size
8742
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 83 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__app(nil, YS) -> mark(YS) a__app(cons(X, XS), YS) -> cons(mark(X), app(XS, YS)) a__from(X) -> cons(mark(X), from(s(X))) a__zWadr(nil, YS) -> nil a__zWadr(XS, nil) -> nil a__zWadr(cons(X, XS), cons(Y, YS)) -> cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS)) a__prefix(L) -> cons(nil, zWadr(L, prefix(L))) mark(app(X1, X2)) -> a__app(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(zWadr(X1, X2)) -> a__zWadr(mark(X1), mark(X2)) mark(prefix(X)) -> a__prefix(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__app(X1, X2) -> app(X1, X2) a__from(X) -> from(X) a__zWadr(X1, X2) -> zWadr(X1, X2) a__prefix(X) -> prefix(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: A__APP(nil, YS) -> MARK(YS) A__APP(cons(X, XS), YS) -> MARK(X) A__FROM(X) -> MARK(X) A__ZWADR(cons(X, XS), cons(Y, YS)) -> A__APP(mark(Y), cons(mark(X), nil)) A__ZWADR(cons(X, XS), cons(Y, YS)) -> MARK(Y) A__ZWADR(cons(X, XS), cons(Y, YS)) -> MARK(X) MARK(app(X1, X2)) -> A__APP(mark(X1), mark(X2)) MARK(app(X1, X2)) -> MARK(X1) MARK(app(X1, X2)) -> MARK(X2) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(zWadr(X1, X2)) -> A__ZWADR(mark(X1), mark(X2)) MARK(zWadr(X1, X2)) -> MARK(X1) MARK(zWadr(X1, X2)) -> MARK(X2) MARK(prefix(X)) -> A__PREFIX(mark(X)) MARK(prefix(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__app(nil, YS) -> mark(YS) a__app(cons(X, XS), YS) -> cons(mark(X), app(XS, YS)) a__from(X) -> cons(mark(X), from(s(X))) a__zWadr(nil, YS) -> nil a__zWadr(XS, nil) -> nil a__zWadr(cons(X, XS), cons(Y, YS)) -> cons(a__app(mark(Y), cons(mark(X), nil)), zWadr(XS, YS)) a__prefix(L) -> cons(nil, zWadr(L, prefix(L))) mark(app(X1, X2)) -> a__app(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(zWadr(X1, X2)) -> a__zWadr(mark(X1), mark(X2)) mark(prefix(X)) -> a__prefix(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__app(X1, X2) -> app(X1, X2) a__from(X) -> from(X)
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