details

property	value
status	complete
benchmark	Ex2_Luc03b_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n026.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.46721887589 seconds
cpu usage	6.490960796
max memory	4.42830848E8

stage attributes

key	value
output-size	11006
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) QDPOrderProof [EQUIVALENT, 74 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 52 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: fst(0, Z) -> nil fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) from(X) -> cons(X, n__from(n__s(X))) add(0, X) -> X add(s(X), Y) -> s(n__add(activate(X), Y)) len(nil) -> 0 len(cons(X, Z)) -> s(n__len(activate(Z))) fst(X1, X2) -> n__fst(X1, X2) from(X) -> n__from(X) s(X) -> n__s(X) add(X1, X2) -> n__add(X1, X2) len(X) -> n__len(X) activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2)) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(X) activate(n__add(X1, X2)) -> add(activate(X1), activate(X2)) activate(n__len(X)) -> len(activate(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: FST(s(X), cons(Y, Z)) -> ACTIVATE(X) FST(s(X), cons(Y, Z)) -> ACTIVATE(Z) ADD(s(X), Y) -> S(n__add(activate(X), Y)) ADD(s(X), Y) -> ACTIVATE(X) LEN(cons(X, Z)) -> S(n__len(activate(Z))) LEN(cons(X, Z)) -> ACTIVATE(Z) ACTIVATE(n__fst(X1, X2)) -> FST(activate(X1), activate(X2)) ACTIVATE(n__fst(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__fst(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__from(X)) -> FROM(activate(X)) ACTIVATE(n__from(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__len(X)) -> LEN(activate(X)) ACTIVATE(n__len(X)) -> ACTIVATE(X) The TRS R consists of the following rules: fst(0, Z) -> nil fst(s(X), cons(Y, Z)) -> cons(Y, n__fst(activate(X), activate(Z))) from(X) -> cons(X, n__from(n__s(X))) add(0, X) -> X add(s(X), Y) -> s(n__add(activate(X), Y)) len(nil) -> 0 len(cons(X, Z)) -> s(n__len(activate(Z))) fst(X1, X2) -> n__fst(X1, X2) from(X) -> n__from(X) s(X) -> n__s(X) add(X1, X2) -> n__add(X1, X2) len(X) -> n__len(X) activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2)) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(X) popout

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

job log

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actions

all output return to TRS Stand 20472