details

property	value
status	complete
benchmark	Ex15_Luc98_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n169.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.00724196434 seconds
cpu usage	8.579058771
max memory	5.86027008E8

stage attributes

key	value
output-size	10604
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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, 134 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: and(true, X) -> activate(X) and(false, Y) -> false if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) add(0, X) -> activate(X) add(s(X), Y) -> s(n__add(activate(X), activate(Y))) first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(activate(Y), n__first(activate(X), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) add(X1, X2) -> n__add(X1, X2) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__add(X1, X2)) -> add(activate(X1), X2) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) activate(n__from(X)) -> from(X) activate(n__s(X)) -> s(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: AND(true, X) -> ACTIVATE(X) IF(true, X, Y) -> ACTIVATE(X) IF(false, X, Y) -> ACTIVATE(Y) ADD(0, X) -> ACTIVATE(X) ADD(s(X), Y) -> S(n__add(activate(X), activate(Y))) ADD(s(X), Y) -> ACTIVATE(X) ADD(s(X), Y) -> ACTIVATE(Y) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z) FROM(X) -> ACTIVATE(X) ACTIVATE(n__add(X1, X2)) -> ADD(activate(X1), X2) ACTIVATE(n__add(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__first(X1, X2)) -> FIRST(activate(X1), activate(X2)) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__from(X)) -> FROM(X) ACTIVATE(n__s(X)) -> S(X) The TRS R consists of the following rules: and(true, X) -> activate(X) and(false, Y) -> false if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) add(0, X) -> activate(X) add(s(X), Y) -> s(n__add(activate(X), activate(Y))) first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(activate(Y), n__first(activate(X), activate(Z))) from(X) -> cons(activate(X), n__from(n__s(activate(X)))) add(X1, X2) -> n__add(X1, X2) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) popout

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

job log

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actions

all output return to TRS Standard