details

property	value
status	complete
benchmark	Ex6_Luc98_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n182.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.24034 seconds
cpu usage	5.3892
user time	5.15386
system time	0.235339
max virtual memory	1.8345424E7
max residence set size	363612.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 4 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 54 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 26 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(n__s(X))) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 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: FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z) 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(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: first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(n__s(X))) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) s(X) -> n__s(X) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) 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 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__first(X1, X2)) -> FIRST(activate(X1), activate(X2)) FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__from(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> ACTIVATE(X) The TRS R consists of the following rules: first(0, X) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) from(X) -> cons(X, n__from(n__s(X))) first(X1, X2) -> n__first(X1, X2) from(X) -> n__from(X) popout

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actions

all output return to TRS Standard