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	n156.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.88669085503 seconds
cpu usage	8.735939249
max memory	5.45181696E8

stage attributes

key	value
output-size	15234
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 19 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 119 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 52 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 40 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) 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) The set Q consists of the following terms: a__from(x0) mark(2nd(x0)) mark(from(x0)) mark(cons(x0, x1)) mark(s(x0)) mark(cons1(x0, x1)) a__2nd(x0) ---------------------------------------- (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) popout

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

job log

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actions

all output return to TRS Innermost