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SRS Stand 10685 pair #381718614
details
property
value
status
complete
benchmark
dup17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n100.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
6.35861301422 seconds
cpu usage
20.899469548
max memory
2.698108928E9
stage attributes
key
value
output-size
8250
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 25 ms] (2) QDP (3) MRRProof [EQUIVALENT, 194 ms] (4) QDP (5) MRRProof [EQUIVALENT, 117 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 1 ms] (8) QDP (9) MRRProof [EQUIVALENT, 31 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(b(x1)) b(b(a(a(x1)))) -> a(a(b(b(x1)))) b(b(b(b(c(c(x1)))))) -> c(c(a(a(x1)))) b(b(b(b(x1)))) -> a(a(a(a(a(a(x1)))))) c(c(a(a(x1)))) -> b(b(a(a(c(c(x1)))))) 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: A(a(a(a(x1)))) -> B(b(x1)) A(a(a(a(x1)))) -> B(x1) B(b(a(a(x1)))) -> A(a(b(b(x1)))) B(b(a(a(x1)))) -> A(b(b(x1))) B(b(a(a(x1)))) -> B(b(x1)) B(b(a(a(x1)))) -> B(x1) B(b(b(b(c(c(x1)))))) -> C(c(a(a(x1)))) B(b(b(b(c(c(x1)))))) -> C(a(a(x1))) B(b(b(b(c(c(x1)))))) -> A(a(x1)) B(b(b(b(c(c(x1)))))) -> A(x1) B(b(b(b(x1)))) -> A(a(a(a(a(a(x1)))))) B(b(b(b(x1)))) -> A(a(a(a(a(x1))))) B(b(b(b(x1)))) -> A(a(a(a(x1)))) B(b(b(b(x1)))) -> A(a(a(x1))) B(b(b(b(x1)))) -> A(a(x1)) B(b(b(b(x1)))) -> A(x1) C(c(a(a(x1)))) -> B(b(a(a(c(c(x1)))))) C(c(a(a(x1)))) -> B(a(a(c(c(x1))))) C(c(a(a(x1)))) -> A(a(c(c(x1)))) C(c(a(a(x1)))) -> A(c(c(x1))) C(c(a(a(x1)))) -> C(c(x1)) C(c(a(a(x1)))) -> C(x1) The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(b(x1)) b(b(a(a(x1)))) -> a(a(b(b(x1)))) b(b(b(b(c(c(x1)))))) -> c(c(a(a(x1)))) b(b(b(b(x1)))) -> a(a(a(a(a(a(x1)))))) c(c(a(a(x1)))) -> b(b(a(a(c(c(x1)))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: B(b(b(b(c(c(x1)))))) -> C(a(a(x1))) B(b(b(b(c(c(x1)))))) -> A(a(x1)) B(b(b(b(c(c(x1)))))) -> A(x1) C(c(a(a(x1)))) -> C(x1)
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