details

property	value
status	complete
benchmark	ExAppendixB_AEL03_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n177.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.90079 seconds
cpu usage	4.19624
user time	4.01538
system time	0.180864
max virtual memory	1.8343148E7
max residence set size	259844.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 16 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) QDPSizeChangeProof [EQUIVALENT, 0 ms] (17) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) 2ndspos(0, Z) -> rnil 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z))) 2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z))) 2ndsneg(0, Z) -> rnil 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z))) 2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, activate(Z))) pi(X) -> 2ndspos(X, from(0)) plus(0, Y) -> Y plus(s(X), Y) -> s(plus(X, Y)) times(0, Y) -> 0 times(s(X), Y) -> plus(Y, times(X, Y)) square(X) -> times(X, X) from(X) -> n__from(X) activate(n__from(X)) -> from(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: 2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z))) 2NDSPOS(s(N), cons(X, Z)) -> ACTIVATE(Z) 2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z)) 2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z) 2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z))) 2NDSNEG(s(N), cons(X, Z)) -> ACTIVATE(Z) 2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z)) 2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z) PI(X) -> 2NDSPOS(X, from(0)) PI(X) -> FROM(0) PLUS(s(X), Y) -> PLUS(X, Y) TIMES(s(X), Y) -> PLUS(Y, times(X, Y)) TIMES(s(X), Y) -> TIMES(X, Y) SQUARE(X) -> TIMES(X, X) ACTIVATE(n__from(X)) -> FROM(X) The TRS R consists of the following rules: from(X) -> cons(X, n__from(s(X))) 2ndspos(0, Z) -> rnil 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z))) 2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z))) 2ndsneg(0, Z) -> rnil 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z))) 2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, activate(Z))) pi(X) -> 2ndspos(X, from(0)) plus(0, Y) -> Y plus(s(X), Y) -> s(plus(X, Y)) times(0, Y) -> 0 times(s(X), Y) -> plus(Y, times(X, Y)) square(X) -> times(X, X) from(X) -> n__from(X) activate(n__from(X)) -> from(X) activate(X) -> X Q is empty. popout

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job log

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actions

all output return to TRS Standard