details

property	value
status	complete
benchmark	Ex6_9_Luc02c_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n178.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	10.65625 seconds
cpu usage	24.479637949
max memory	1.22370048E9

stage attributes

key	value
output-size	13322
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 90 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 63 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 22 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__2nd(cons1(X, cons(Y, Z))) -> mark(Y) a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1))) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2)) a__2nd(X) -> 2nd(X) a__from(X) -> from(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: A__2ND(cons1(X, cons(Y, Z))) -> MARK(Y) A__2ND(cons(X, X1)) -> A__2ND(cons1(mark(X), mark(X1))) A__2ND(cons(X, X1)) -> MARK(X) A__2ND(cons(X, X1)) -> MARK(X1) A__FROM(X) -> MARK(X) MARK(2nd(X)) -> A__2ND(mark(X)) MARK(2nd(X)) -> MARK(X) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) MARK(cons1(X1, X2)) -> MARK(X1) MARK(cons1(X1, X2)) -> MARK(X2) The TRS R consists of the following rules: a__2nd(cons1(X, cons(Y, Z))) -> mark(Y) a__2nd(cons(X, X1)) -> a__2nd(cons1(mark(X), mark(X1))) a__from(X) -> cons(mark(X), from(s(X))) mark(2nd(X)) -> a__2nd(mark(X)) mark(from(X)) -> a__from(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) mark(cons1(X1, X2)) -> cons1(mark(X1), mark(X2)) a__2nd(X) -> 2nd(X) a__from(X) -> from(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. popout

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all output return to TRS Standard