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	n118.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	4.57848596573 seconds
cpu usage	14.696649664
max memory	9.53475072E8

stage attributes

key	value
output-size	20689
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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, 217 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 155 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 50 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) popout

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

job log

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actions

all output return to TRS Standard