details

property	value
status	complete
benchmark	t012.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74338 seconds
cpu usage	3.87445
user time	3.71647
system time	0.157979
max virtual memory	1.8409712E7
max residence set size	246852.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 30 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) MRRProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(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(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) MINUS(*(x, y)) -> MINUS(minus(x)) MINUS(*(x, y)) -> MINUS(x) MINUS(*(x, y)) -> MINUS(minus(minus(y))) MINUS(*(x, y)) -> MINUS(minus(y)) MINUS(*(x, y)) -> MINUS(y) F(minus(x)) -> MINUS(minus(minus(f(x)))) F(minus(x)) -> MINUS(minus(f(x))) F(minus(x)) -> MINUS(f(x)) F(minus(x)) -> F(x) The TRS R consists of the following rules: minus(minus(x)) -> x minus(+(x, y)) -> *(minus(minus(minus(x))), minus(minus(minus(y)))) minus(*(x, y)) -> +(minus(minus(minus(x))), minus(minus(minus(y)))) f(minus(x)) -> minus(minus(minus(f(x)))) 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 3 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(+(x, y)) -> MINUS(minus(x)) MINUS(+(x, y)) -> MINUS(minus(minus(x))) MINUS(+(x, y)) -> MINUS(x) MINUS(+(x, y)) -> MINUS(minus(minus(y))) MINUS(+(x, y)) -> MINUS(minus(y)) MINUS(+(x, y)) -> MINUS(y) MINUS(*(x, y)) -> MINUS(minus(minus(x))) popout

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