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TRS Stand 20472 pair #381713865
details
property
value
status
complete
benchmark
gen-14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n108.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.97289609909 seconds
cpu usage
4.799346895
max memory
3.15805696E8
stage attributes
key
value
output-size
4075
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 21 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 29 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(f(b(x, z)), y) -> f(f(f(b(z, b(y, z))))) c(f(f(c(x, a, z))), a, y) -> b(y, f(b(a, z))) b(b(c(b(a, a), a, z), f(a)), y) -> 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(f(b(x, z)), y) -> B(z, b(y, z)) B(f(b(x, z)), y) -> B(y, z) C(f(f(c(x, a, z))), a, y) -> B(y, f(b(a, z))) C(f(f(c(x, a, z))), a, y) -> B(a, z) The TRS R consists of the following rules: b(f(b(x, z)), y) -> f(f(f(b(z, b(y, z))))) c(f(f(c(x, a, z))), a, y) -> b(y, f(b(a, z))) b(b(c(b(a, a), a, z), f(a)), y) -> 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(f(b(x, z)), y) -> B(y, z) B(f(b(x, z)), y) -> B(z, b(y, z)) The TRS R consists of the following rules: b(f(b(x, z)), y) -> f(f(f(b(z, b(y, z))))) c(f(f(c(x, a, z))), a, y) -> b(y, f(b(a, z))) b(b(c(b(a, a), a, z), f(a)), y) -> 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. B(f(b(x, z)), y) -> B(y, z) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) : <<< POL(B(x_1, x_2)) = [[1]] + [[0, 1]] * x_1 + [[0, 1]] * x_2 >>> <<<
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