details

property	value
status	complete
benchmark	Ex6_9_Luc02c.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.07597899437 seconds
cpu usage	4.916796362
max memory	3.73039104E8

stage attributes

key	value
output-size	5643
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) NonTerminationLoopProof [COMPLETE, 0 ms] (16) NO ---------------------------------------- (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, X1)) from(X) -> cons(X, from(s(X))) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) 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, X1)) from(X) -> cons(X, from(s(X))) The set Q consists of the following terms: 2nd(cons1(x0, cons(x1, x2))) 2nd(cons(x0, x1)) from(x0) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 2ND(cons(X, X1)) -> 2ND(cons1(X, X1)) FROM(X) -> FROM(s(X)) The TRS R consists of the following rules: 2nd(cons1(X, cons(Y, Z))) -> Y 2nd(cons(X, X1)) -> 2nd(cons1(X, X1)) from(X) -> cons(X, from(s(X))) The set Q consists of the following terms: 2nd(cons1(x0, cons(x1, x2))) 2nd(cons(x0, x1)) from(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- popout

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job log

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actions

all output return to TRS Stand 20472