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SRS Stand 10685 pair #381719220
details
property
value
status
complete
benchmark
z125.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n105.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.41923284531 seconds
cpu usage
10.470101182
max memory
1.630593024E9
stage attributes
key
value
output-size
7625
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 1 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 20 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 7 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x1) -> n(c(n(a(x1)))) c(f(x1)) -> f(n(a(c(x1)))) n(a(x1)) -> c(x1) c(c(x1)) -> c(x1) n(s(x1)) -> f(s(s(x1))) n(f(x1)) -> f(n(x1)) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x1) -> a(n(c(n(x1)))) f(c(x1)) -> c(a(n(f(x1)))) a(n(x1)) -> c(x1) c(c(x1)) -> c(x1) s(n(x1)) -> s(s(f(x1))) f(n(x1)) -> n(f(x1)) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(x1) -> A(n(c(n(x1)))) F(x1) -> C(n(x1)) F(c(x1)) -> C(a(n(f(x1)))) F(c(x1)) -> A(n(f(x1))) F(c(x1)) -> F(x1) A(n(x1)) -> C(x1) S(n(x1)) -> S(s(f(x1))) S(n(x1)) -> S(f(x1)) S(n(x1)) -> F(x1) F(n(x1)) -> F(x1) The TRS R consists of the following rules: f(x1) -> a(n(c(n(x1)))) f(c(x1)) -> c(a(n(f(x1)))) a(n(x1)) -> c(x1) c(c(x1)) -> c(x1)
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