details

property	value
status	complete
benchmark	cime5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n102.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.55069208145 seconds
cpu usage	16.592386175
max memory	1.600274432E9

stage attributes

key	value
output-size	40691
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 61 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 579 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 599 ms] (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (20) QDP (21) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: intersect'ii'in(cons(X, X0), cons(X, X1)) -> intersect'ii'out intersect'ii'in(Xs, cons(X0, Ys)) -> u'1'1(intersect'ii'in(Xs, Ys)) u'1'1(intersect'ii'out) -> intersect'ii'out intersect'ii'in(cons(X0, Xs), Ys) -> u'2'1(intersect'ii'in(Xs, Ys)) u'2'1(intersect'ii'out) -> intersect'ii'out reduce'ii'in(sequent(cons(if(A, B), Fs), Gs), NF) -> u'3'1(reduce'ii'in(sequent(cons(x'2b(x'2d(B), A), Fs), Gs), NF)) u'3'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(cons(iff(A, B), Fs), Gs), NF) -> u'4'1(reduce'ii'in(sequent(cons(x'2a(if(A, B), if(B, A)), Fs), Gs), NF)) u'4'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(cons(x'2a(F1, F2), Fs), Gs), NF) -> u'5'1(reduce'ii'in(sequent(cons(F1, cons(F2, Fs)), Gs), NF)) u'5'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(cons(x'2b(F1, F2), Fs), Gs), NF) -> u'6'1(reduce'ii'in(sequent(cons(F1, Fs), Gs), NF), F2, Fs, Gs, NF) u'6'1(reduce'ii'out, F2, Fs, Gs, NF) -> u'6'2(reduce'ii'in(sequent(cons(F2, Fs), Gs), NF)) u'6'2(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(cons(x'2d(F1), Fs), Gs), NF) -> u'7'1(reduce'ii'in(sequent(Fs, cons(F1, Gs)), NF)) u'7'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(Fs, cons(if(A, B), Gs)), NF) -> u'8'1(reduce'ii'in(sequent(Fs, cons(x'2b(x'2d(B), A), Gs)), NF)) u'8'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(Fs, cons(iff(A, B), Gs)), NF) -> u'9'1(reduce'ii'in(sequent(Fs, cons(x'2a(if(A, B), if(B, A)), Gs)), NF)) u'9'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(cons(p(V), Fs), Gs), sequent(Left, Right)) -> u'10'1(reduce'ii'in(sequent(Fs, Gs), sequent(cons(p(V), Left), Right))) u'10'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(Fs, cons(x'2b(G1, G2), Gs)), NF) -> u'11'1(reduce'ii'in(sequent(Fs, cons(G1, cons(G2, Gs))), NF)) u'11'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(Fs, cons(x'2a(G1, G2), Gs)), NF) -> u'12'1(reduce'ii'in(sequent(Fs, cons(G1, Gs)), NF), Fs, G2, Gs, NF) u'12'1(reduce'ii'out, Fs, G2, Gs, NF) -> u'12'2(reduce'ii'in(sequent(Fs, cons(G2, Gs)), NF)) u'12'2(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(Fs, cons(x'2d(G1), Gs)), NF) -> u'13'1(reduce'ii'in(sequent(cons(G1, Fs), Gs), NF)) u'13'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(nil, cons(p(V), Gs)), sequent(Left, Right)) -> u'14'1(reduce'ii'in(sequent(nil, Gs), sequent(Left, cons(p(V), Right)))) u'14'1(reduce'ii'out) -> reduce'ii'out reduce'ii'in(sequent(nil, nil), sequent(F1, F2)) -> u'15'1(intersect'ii'in(F1, F2)) u'15'1(intersect'ii'out) -> reduce'ii'out tautology'i'in(F) -> u'16'1(reduce'ii'in(sequent(nil, cons(F, nil)), sequent(nil, nil))) u'16'1(reduce'ii'out) -> tautology'i'out 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: INTERSECT'II'IN(Xs, cons(X0, Ys)) -> U'1'1(intersect'ii'in(Xs, Ys)) INTERSECT'II'IN(Xs, cons(X0, Ys)) -> INTERSECT'II'IN(Xs, Ys) INTERSECT'II'IN(cons(X0, Xs), Ys) -> U'2'1(intersect'ii'in(Xs, Ys)) INTERSECT'II'IN(cons(X0, Xs), Ys) -> INTERSECT'II'IN(Xs, Ys) popout

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actions

all output return to TRS Stand 20472