details

property	value
status	complete
benchmark	Ex6_9_Luc02c_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n098.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.78880596161 seconds
cpu usage	3.697003265
max memory	1.93691648E8

stage attributes

key	value
output-size	3862
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) from(X) -> cons(X, n__from(n__s(X))) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(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: 2ND(cons(X, X1)) -> 2ND(cons1(X, activate(X1))) 2ND(cons(X, X1)) -> ACTIVATE(X1) ACTIVATE(n__from(X)) -> FROM(activate(X)) ACTIVATE(n__from(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> S(activate(X)) ACTIVATE(n__s(X)) -> ACTIVATE(X) The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) from(X) -> cons(X, n__from(n__s(X))) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__s(X)) -> ACTIVATE(X) ACTIVATE(n__from(X)) -> ACTIVATE(X) The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) from(X) -> cons(X, n__from(n__s(X))) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. popout

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job log

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actions

all output return to TRS Stand 20472