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TRS Stand 20472 pair #381713054
details
property
value
status
complete
benchmark
gen-22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n048.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.34245300293 seconds
cpu usage
5.904948148
max memory
3.73940224E8
stage attributes
key
value
output-size
6941
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 3 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 9 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a, b(c(z, x, y), a)) -> b(b(z, c(y, z, a)), x) f(c(a, b(b(z, a), y), x)) -> f(c(x, b(z, x), y)) c(f(c(a, y, a)), x, z) -> f(b(b(z, z), f(b(y, b(x, a))))) 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(a, b(c(z, x, y), a)) -> B(b(z, c(y, z, a)), x) B(a, b(c(z, x, y), a)) -> B(z, c(y, z, a)) B(a, b(c(z, x, y), a)) -> C(y, z, a) F(c(a, b(b(z, a), y), x)) -> F(c(x, b(z, x), y)) F(c(a, b(b(z, a), y), x)) -> C(x, b(z, x), y) F(c(a, b(b(z, a), y), x)) -> B(z, x) C(f(c(a, y, a)), x, z) -> F(b(b(z, z), f(b(y, b(x, a))))) C(f(c(a, y, a)), x, z) -> B(b(z, z), f(b(y, b(x, a)))) C(f(c(a, y, a)), x, z) -> B(z, z) C(f(c(a, y, a)), x, z) -> F(b(y, b(x, a))) C(f(c(a, y, a)), x, z) -> B(y, b(x, a)) C(f(c(a, y, a)), x, z) -> B(x, a) The TRS R consists of the following rules: b(a, b(c(z, x, y), a)) -> b(b(z, c(y, z, a)), x) f(c(a, b(b(z, a), y), x)) -> f(c(x, b(z, x), y)) c(f(c(a, y, a)), x, z) -> f(b(b(z, z), f(b(y, b(x, a))))) 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 2 SCCs with 9 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: B(a, b(c(z, x, y), a)) -> C(y, z, a) C(f(c(a, y, a)), x, z) -> B(y, b(x, a)) The TRS R consists of the following rules: b(a, b(c(z, x, y), a)) -> b(b(z, c(y, z, a)), x) f(c(a, b(b(z, a), y), x)) -> f(c(x, b(z, x), y))
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