details

property	value
status	complete
benchmark	test77.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.53678 seconds
cpu usage	11.74
user time	11.0459
system time	0.694106
max virtual memory	1.841326E7
max residence set size	2323272.0

stage attributes

key	value
starexec-result	NO

output

NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) NonTerminationLoopProof [COMPLETE, 3490 ms] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: +(X, 0) -> X +(X, s(Y)) -> s(+(X, Y)) double(X) -> +(X, X) f(0, s(0), X) -> f(X, double(X), X) g(X, Y) -> X g(X, Y) -> Y 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: +^1(X, s(Y)) -> +^1(X, Y) DOUBLE(X) -> +^1(X, X) F(0, s(0), X) -> F(X, double(X), X) F(0, s(0), X) -> DOUBLE(X) The TRS R consists of the following rules: +(X, 0) -> X +(X, s(Y)) -> s(+(X, Y)) double(X) -> +(X, X) f(0, s(0), X) -> f(X, double(X), X) g(X, Y) -> X g(X, Y) -> Y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: +^1(X, s(Y)) -> +^1(X, Y) The TRS R consists of the following rules: +(X, 0) -> X +(X, s(Y)) -> s(+(X, Y)) double(X) -> +(X, X) f(0, s(0), X) -> f(X, double(X), X) g(X, Y) -> X g(X, Y) -> Y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- popout

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all output return to TRS Standard