details

property	value
status	complete
benchmark	Ex49_GM04_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.82643 seconds
cpu usage	11.346
user time	10.8443
system time	0.501702
max virtual memory	3.6850556E7
max residence set size	784532.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 193 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 112 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 96 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(n__0, Y) -> 0 minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y)) geq(X, n__0) -> true geq(n__0, n__s(Y)) -> false geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y)) div(0, n__s(Y)) -> 0 div(s(X), n__s(Y)) -> if(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) 0 -> n__0 s(X) -> n__s(X) div(X1, X2) -> n__div(X1, X2) minus(X1, X2) -> n__minus(X1, X2) activate(n__0) -> 0 activate(n__s(X)) -> s(activate(X)) activate(n__div(X1, X2)) -> div(activate(X1), X2) activate(n__minus(X1, X2)) -> minus(X1, X2) 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: MINUS(n__0, Y) -> 0^1 MINUS(n__s(X), n__s(Y)) -> MINUS(activate(X), activate(Y)) MINUS(n__s(X), n__s(Y)) -> ACTIVATE(X) MINUS(n__s(X), n__s(Y)) -> ACTIVATE(Y) GEQ(n__s(X), n__s(Y)) -> GEQ(activate(X), activate(Y)) GEQ(n__s(X), n__s(Y)) -> ACTIVATE(X) GEQ(n__s(X), n__s(Y)) -> ACTIVATE(Y) DIV(s(X), n__s(Y)) -> IF(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0) DIV(s(X), n__s(Y)) -> GEQ(X, activate(Y)) DIV(s(X), n__s(Y)) -> ACTIVATE(Y) IF(true, X, Y) -> ACTIVATE(X) IF(false, X, Y) -> ACTIVATE(Y) ACTIVATE(n__0) -> 0^1 ACTIVATE(n__s(X)) -> S(activate(X)) ACTIVATE(n__s(X)) -> ACTIVATE(X) ACTIVATE(n__div(X1, X2)) -> DIV(activate(X1), X2) ACTIVATE(n__div(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__minus(X1, X2)) -> MINUS(X1, X2) The TRS R consists of the following rules: minus(n__0, Y) -> 0 minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y)) geq(X, n__0) -> true geq(n__0, n__s(Y)) -> false geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y)) div(0, n__s(Y)) -> 0 div(s(X), n__s(Y)) -> if(geq(X, activate(Y)), n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))), n__0) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) 0 -> n__0 s(X) -> n__s(X) div(X1, X2) -> n__div(X1, X2) minus(X1, X2) -> n__minus(X1, X2) activate(n__0) -> 0 popout

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

job log

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actions

all output return to TRS Standard