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	n081.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.78339600563 seconds
cpu usage	5.779230732
max memory	4.66259968E8

stage attributes

key	value
output-size	7593
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 38 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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)) popout

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all output return to TRS Standard