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TRS Relative pair #487520754
details
property
value
status
complete
benchmark
rt1-3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n143.star.cs.uiowa.edu
space
Relative_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.7118601799 seconds
cpu usage
3.578184196
max memory
2.209792E8
stage attributes
key
value
output-size
2509
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRStoQDPProof [SOUND, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 47 ms] (4) QDP (5) PisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) The relative TRS consists of the following S rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(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: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) The TRS R consists of the following rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(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 dependency pairs: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) Used ordering: Polynomial interpretation [POLO]: POL(f(x_1)) = 2 + 2*x_1 POL(g(x_1)) = 2 + 2*x_1 POL(t(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3 ---------------------------------------- (4) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(x)) Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (5) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain.
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