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TRS Standard pair #516966170
details
property
value
status
complete
benchmark
gen-10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n137.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.50121092796 seconds
cpu usage
6.242359995
max memory
4.8877568E8
stage attributes
key
value
output-size
5076
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(b(a, z)) -> z b(y, b(a, z)) -> b(f(c(y, y, a)), b(f(z), a)) f(f(f(c(z, x, a)))) -> b(f(x), z) 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: B(y, b(a, z)) -> B(f(c(y, y, a)), b(f(z), a)) B(y, b(a, z)) -> F(c(y, y, a)) B(y, b(a, z)) -> B(f(z), a) B(y, b(a, z)) -> F(z) F(f(f(c(z, x, a)))) -> B(f(x), z) F(f(f(c(z, x, a)))) -> F(x) The TRS R consists of the following rules: f(b(a, z)) -> z b(y, b(a, z)) -> b(f(c(y, y, a)), b(f(z), a)) f(f(f(c(z, x, a)))) -> b(f(x), z) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(y, b(a, z)) -> F(z) F(f(f(c(z, x, a)))) -> B(f(x), z) B(y, b(a, z)) -> B(f(c(y, y, a)), b(f(z), a)) F(f(f(c(z, x, a)))) -> F(x) The TRS R consists of the following rules: f(b(a, z)) -> z b(y, b(a, z)) -> b(f(c(y, y, a)), b(f(z), a)) f(f(f(c(z, x, a)))) -> b(f(x), z) 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. F(f(f(c(z, x, a)))) -> B(f(x), z)
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