details

property	value
status	complete
benchmark	ExAppendixB_AEL03_FR.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)	1.96192 seconds
cpu usage	4.28741
user time	4.08592
system time	0.201489
max virtual memory	1.8408912E7
max residence set size	267348.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: from(X) -> cons(X, n__from(n__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) s(X) -> n__s(X) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(activate(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) -> S(plus(X, Y)) 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(activate(X)) ACTIVATE(n__from(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> S(activate(X)) ACTIVATE(n__s(X)) -> ACTIVATE(X) The TRS R consists of the following rules: from(X) -> cons(X, n__from(n__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))) popout

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

job log

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actions

all output return to TRS Standard