details

property	value
status	complete
benchmark	pair3swap.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n116.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.51958012581 seconds
cpu usage	7.645497994
max memory	5.14609152E8

stage attributes

key	value
output-size	3709
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 36 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 53 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(a(a(x0)), p(x1, p(a(x2), x3))) -> p(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) 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: P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(a(a(b(x1))), p(a(a(x0)), x3)) P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(a(a(x0)), x3) The TRS R consists of the following rules: p(a(a(x0)), p(x1, p(a(x2), x3))) -> p(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(a(a(b(x1))), p(a(a(x0)), x3)) P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(a(a(x0)), x3) Used ordering: Polynomial interpretation [POLO]: POL(P(x_1, x_2)) = 2*x_1 + x_2 POL(a(x_1)) = x_1 POL(b(x_1)) = x_1 POL(p(x_1, x_2)) = 1 + 2*x_1 + x_2 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) The TRS R consists of the following rules: p(a(a(x0)), p(x1, p(a(x2), x3))) -> p(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. P(a(a(x0)), p(x1, p(a(x2), x3))) -> P(x2, p(a(a(b(x1))), p(a(a(x0)), x3))) The remaining pairs can at least be oriented weakly. popout

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