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SRS Relative pair #487520971
details
property
value
status
complete
benchmark
dup02.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
Mixed_relative_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
23.2186629772 seconds
cpu usage
88.699125102
max memory
6.104989696E9
stage attributes
key
value
output-size
11960
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRS Reverse [EQUIVALENT, 0 ms] (2) RelTRS (3) RelTRSRRRProof [EQUIVALENT, 70 ms] (4) RelTRS (5) RelTRSRRRProof [EQUIVALENT, 3078 ms] (6) RelTRS (7) RelTRSRRRProof [EQUIVALENT, 5 ms] (8) RelTRS (9) RelTRSRRRProof [EQUIVALENT, 545 ms] (10) RelTRS (11) RelTRSRRRProof [EQUIVALENT, 319 ms] (12) RelTRS (13) RelTRSRRRProof [EQUIVALENT, 964 ms] (14) RelTRS (15) RIsEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: d(d(n(n(x1)))) -> d(d(x1)) d(d(o(o(x1)))) -> d(d(x1)) o(o(u(u(x1)))) -> u(u(x1)) The relative TRS consists of the following S rules: t(t(u(u(x1)))) -> t(t(c(c(d(d(x1)))))) d(d(f(f(x1)))) -> f(f(d(d(x1)))) d(d(g(g(x1)))) -> u(u(g(g(x1)))) f(f(u(u(x1)))) -> u(u(f(f(x1)))) n(n(u(u(x1)))) -> u(u(x1)) f(f(x1)) -> f(f(n(n(x1)))) t(t(x1)) -> t(t(c(c(n(n(x1)))))) c(c(n(n(x1)))) -> n(n(c(c(x1)))) c(c(o(o(x1)))) -> o(o(c(c(x1)))) c(c(o(o(x1)))) -> o(o(x1)) c(c(f(f(x1)))) -> f(f(c(c(x1)))) c(c(u(u(x1)))) -> u(u(c(c(x1)))) c(c(d(d(x1)))) -> d(d(c(c(x1)))) ---------------------------------------- (1) RelTRS Reverse (EQUIVALENT) We have reversed the following relative TRS [REVERSE]: The set of rules R is d(d(n(n(x1)))) -> d(d(x1)) d(d(o(o(x1)))) -> d(d(x1)) o(o(u(u(x1)))) -> u(u(x1)) The set of rules S is t(t(u(u(x1)))) -> t(t(c(c(d(d(x1)))))) d(d(f(f(x1)))) -> f(f(d(d(x1)))) d(d(g(g(x1)))) -> u(u(g(g(x1)))) f(f(u(u(x1)))) -> u(u(f(f(x1)))) n(n(u(u(x1)))) -> u(u(x1)) f(f(x1)) -> f(f(n(n(x1)))) t(t(x1)) -> t(t(c(c(n(n(x1)))))) c(c(n(n(x1)))) -> n(n(c(c(x1)))) c(c(o(o(x1)))) -> o(o(c(c(x1)))) c(c(o(o(x1)))) -> o(o(x1)) c(c(f(f(x1)))) -> f(f(c(c(x1)))) c(c(u(u(x1)))) -> u(u(c(c(x1)))) c(c(d(d(x1)))) -> d(d(c(c(x1)))) We have obtained the following relative TRS: The set of rules R is n(n(d(d(x1)))) -> d(d(x1)) o(o(d(d(x1)))) -> d(d(x1)) u(u(o(o(x1)))) -> u(u(x1)) The set of rules S is u(u(t(t(x1)))) -> d(d(c(c(t(t(x1)))))) f(f(d(d(x1)))) -> d(d(f(f(x1)))) g(g(d(d(x1)))) -> g(g(u(u(x1)))) u(u(f(f(x1)))) -> f(f(u(u(x1)))) u(u(n(n(x1)))) -> u(u(x1)) f(f(x1)) -> n(n(f(f(x1)))) t(t(x1)) -> n(n(c(c(t(t(x1)))))) n(n(c(c(x1)))) -> c(c(n(n(x1)))) o(o(c(c(x1)))) -> c(c(o(o(x1))))
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