details

property	value
status	complete
benchmark	#3.49_rand.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n107.star.cs.uiowa.edu
space	INVY_15

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.97083 seconds
cpu usage	4.26596
user time	4.08792
system time	0.178041
max virtual memory	1.8610856E7
max residence set size	277384.0

stage attributes

key	value
starexec-result	YES

output

4.17/1.93 YES 4.17/1.94 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 4.17/1.94 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.17/1.94 4.17/1.94 4.17/1.94 Termination of the given RelTRS could be proven: 4.17/1.94 4.17/1.94 (0) RelTRS 4.17/1.94 (1) RelTRStoQDPProof [SOUND, 0 ms] 4.17/1.94 (2) QDP 4.17/1.94 (3) MRRProof [EQUIVALENT, 53 ms] 4.17/1.94 (4) QDP 4.17/1.94 (5) QDPOrderProof [EQUIVALENT, 36 ms] 4.17/1.94 (6) QDP 4.17/1.94 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 4.17/1.94 (8) AND 4.17/1.94 (9) QDP 4.17/1.94 (10) QDPOrderProof [EQUIVALENT, 6 ms] 4.17/1.94 (11) QDP 4.17/1.94 (12) PisEmptyProof [EQUIVALENT, 0 ms] 4.17/1.94 (13) YES 4.17/1.94 (14) QDP 4.17/1.94 (15) QDPOrderProof [EQUIVALENT, 0 ms] 4.17/1.94 (16) QDP 4.17/1.94 (17) PisEmptyProof [EQUIVALENT, 0 ms] 4.17/1.94 (18) YES 4.17/1.94 4.17/1.94 4.17/1.94 ---------------------------------------- 4.17/1.94 4.17/1.94 (0) 4.17/1.94 Obligation: 4.17/1.94 Relative term rewrite system: 4.17/1.94 The relative TRS consists of the following R rules: 4.17/1.94 4.17/1.94 f(c(s(x), y)) -> f(c(x, s(y))) 4.17/1.94 f(c(s(x), s(y))) -> g(c(x, y)) 4.17/1.94 g(c(x, s(y))) -> g(c(s(x), y)) 4.17/1.94 g(c(s(x), s(y))) -> f(c(x, y)) 4.17/1.94 4.17/1.94 The relative TRS consists of the following S rules: 4.17/1.94 4.17/1.94 rand(x) -> x 4.17/1.94 rand(x) -> rand(s(x)) 4.17/1.94 4.17/1.94 4.17/1.94 ---------------------------------------- 4.17/1.94 4.17/1.94 (1) RelTRStoQDPProof (SOUND) 4.17/1.94 The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem. 4.17/1.94 ---------------------------------------- 4.17/1.94 4.17/1.94 (2) 4.17/1.94 Obligation: 4.17/1.94 Q DP problem: 4.17/1.94 The TRS P consists of the following rules: 4.17/1.94 4.17/1.94 f(c(s(x), y)) -> f(c(x, s(y))) 4.17/1.94 f(c(s(x), s(y))) -> g(c(x, y)) 4.17/1.94 g(c(x, s(y))) -> g(c(s(x), y)) 4.17/1.94 g(c(s(x), s(y))) -> f(c(x, y)) 4.17/1.94 4.17/1.94 The TRS R consists of the following rules: 4.17/1.94 4.17/1.94 rand(x) -> x 4.17/1.94 rand(x) -> rand(s(x)) 4.17/1.94 4.17/1.94 Q is empty. 4.17/1.94 We have to consider all (P,Q,R)-chains. 4.17/1.94 ---------------------------------------- 4.17/1.94 4.17/1.94 (3) MRRProof (EQUIVALENT) 4.17/1.94 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. 4.17/1.94 4.17/1.94 4.17/1.94 Strictly oriented rules of the TRS R: 4.17/1.94 4.17/1.94 rand(x) -> x 4.17/1.94 4.17/1.94 Used ordering: Polynomial interpretation [POLO]: 4.17/1.94 4.17/1.94 POL(c(x_1, x_2)) = x_1 + 2*x_2 4.17/1.94 POL(f(x_1)) = 2*x_1 4.17/1.94 POL(g(x_1)) = 2*x_1 4.17/1.94 POL(rand(x_1)) = 1 + 2*x_1 4.17/1.94 POL(s(x_1)) = x_1 4.17/1.94 4.17/1.94 4.17/1.94 ---------------------------------------- 4.17/1.94 4.17/1.94 (4) 4.17/1.94 Obligation: 4.17/1.94 Q DP problem: 4.17/1.94 The TRS P consists of the following rules: 4.17/1.94 4.17/1.94 f(c(s(x), y)) -> f(c(x, s(y))) 4.17/1.94 f(c(s(x), s(y))) -> g(c(x, y)) 4.17/1.94 g(c(x, s(y))) -> g(c(s(x), y)) 4.17/1.94 g(c(s(x), s(y))) -> f(c(x, y)) 4.17/1.94 popout

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all output return to TRS_Relative 2019-03-21 03.54