details

property	value
status	complete
benchmark	Ex26_Luc03b_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n181.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.91026 seconds
cpu usage	11.7893
user time	11.2652
system time	0.524184
max virtual memory	1.8492108E7
max residence set size	1002856.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 13 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 187 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 162 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 190 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) sqr(0) -> 0 sqr(s(X)) -> s(n__add(n__sqr(activate(X)), n__dbl(activate(X)))) dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) add(0, X) -> X add(s(X), Y) -> s(n__add(activate(X), Y)) first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(activate(X), activate(Z))) terms(X) -> n__terms(X) s(X) -> n__s(X) add(X1, X2) -> n__add(X1, X2) sqr(X) -> n__sqr(X) dbl(X) -> n__dbl(X) first(X1, X2) -> n__first(X1, X2) activate(n__terms(X)) -> terms(activate(X)) activate(n__s(X)) -> s(X) activate(n__add(X1, X2)) -> add(activate(X1), activate(X2)) activate(n__sqr(X)) -> sqr(activate(X)) activate(n__dbl(X)) -> dbl(activate(X)) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 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: TERMS(N) -> SQR(N) SQR(s(X)) -> S(n__add(n__sqr(activate(X)), n__dbl(activate(X)))) SQR(s(X)) -> ACTIVATE(X) DBL(s(X)) -> S(n__s(n__dbl(activate(X)))) DBL(s(X)) -> ACTIVATE(X) ADD(s(X), Y) -> S(n__add(activate(X), Y)) ADD(s(X), Y) -> ACTIVATE(X) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z) ACTIVATE(n__terms(X)) -> TERMS(activate(X)) ACTIVATE(n__terms(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> S(X) ACTIVATE(n__add(X1, X2)) -> ADD(activate(X1), activate(X2)) ACTIVATE(n__add(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__add(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__sqr(X)) -> SQR(activate(X)) ACTIVATE(n__sqr(X)) -> ACTIVATE(X) ACTIVATE(n__dbl(X)) -> DBL(activate(X)) ACTIVATE(n__dbl(X)) -> ACTIVATE(X) ACTIVATE(n__first(X1, X2)) -> FIRST(activate(X1), activate(X2)) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) sqr(0) -> 0 sqr(s(X)) -> s(n__add(n__sqr(activate(X)), n__dbl(activate(X)))) dbl(0) -> 0 dbl(s(X)) -> s(n__s(n__dbl(activate(X)))) add(0, X) -> X add(s(X), Y) -> s(n__add(activate(X), Y)) first(0, X) -> nil popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard