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	n178.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.21786 seconds
cpu usage	4.97565
user time	4.72286
system time	0.252787
max virtual memory	1.8477752E7
max residence set size	340628.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 19 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 78 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. Used ordering: Polynomial interpretation [POLO,RATPOLO]: POL(P(x_1, x_2)) = [1/4]x_1 + [1/2]x_2 POL(a(x_1)) = [1/4] + [4]x_1 POL(b(x_1)) = 0 POL(p(x_1, x_2)) = [1/4]x_1 + [1/2]x_2 popout

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all output return to TRS Standard