details

property	value
status	complete
benchmark	Ex49_GM04_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n183.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.01996 seconds
cpu usage	4.54424
user time	4.32295
system time	0.221285
max virtual memory	1.868274E7
max residence set size	335600.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) MNOCProof [EQUIVALENT, 0 ms] (9) QDP (10) MRRProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) MNOCProof [EQUIVALENT, 0 ms] (18) QDP (19) MRRProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) QDPOrderProof [EQUIVALENT, 0 ms] (25) QDP (26) PisEmptyProof [EQUIVALENT, 0 ms] (27) YES ---------------------------------------- (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(div(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) activate(n__0) -> 0 activate(n__s(X)) -> s(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: 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(div(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) DIV(s(X), n__s(Y)) -> DIV(minus(X, activate(Y)), n__s(activate(Y))) DIV(s(X), n__s(Y)) -> MINUS(X, activate(Y)) IF(true, X, Y) -> ACTIVATE(X) IF(false, X, Y) -> ACTIVATE(Y) ACTIVATE(n__0) -> 0^1 ACTIVATE(n__s(X)) -> S(X) 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(div(minus(X, activate(Y)), n__s(activate(Y)))), n__0) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) popout

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

job log

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actions

all output return to TRS Standard