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TRS Relative pair #487081725
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
n140.star.cs.uiowa.edu
space
INVY_15
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.22499 seconds
cpu usage
4.875
user time
4.67257
system time
0.202428
max virtual memory
1.8544088E7
max residence set size
323840.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRStoQDPProof [SOUND, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 35 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 32 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 1 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: f(c(s(x), y)) -> f(c(x, s(y))) f(c(s(x), s(y))) -> g(c(x, y)) g(c(x, s(y))) -> g(c(s(x), y)) g(c(s(x), s(y))) -> f(c(x, y)) The relative TRS consists of the following S rules: rand(x) -> x rand(x) -> rand(s(x)) ---------------------------------------- (1) RelTRStoQDPProof (SOUND) The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: f(c(s(x), y)) -> f(c(x, s(y))) f(c(s(x), s(y))) -> g(c(x, y)) g(c(x, s(y))) -> g(c(s(x), y)) g(c(s(x), s(y))) -> f(c(x, y)) The TRS R consists of the following rules: rand(x) -> x rand(x) -> rand(s(x)) Q is empty. We have to consider all (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 rules of the TRS R: rand(x) -> x Used ordering: Polynomial interpretation [POLO]: POL(c(x_1, x_2)) = x_1 + 2*x_2 POL(f(x_1)) = 2*x_1 POL(g(x_1)) = 2*x_1 POL(rand(x_1)) = 1 + 2*x_1 POL(s(x_1)) = x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: f(c(s(x), y)) -> f(c(x, s(y))) f(c(s(x), s(y))) -> g(c(x, y)) g(c(x, s(y))) -> g(c(s(x), y)) g(c(s(x), s(y))) -> f(c(x, y))
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