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SRS Standard pair #487513115
details
property
value
status
complete
benchmark
size-12-alpha-3-num-360.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n131.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.90530204773 seconds
cpu usage
12.111164955
max memory
1.29312768E9
stage attributes
key
value
output-size
4419
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 16 ms] (4) QDP (5) MRRProof [EQUIVALENT, 52 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 124 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(x1) a(b(b(x1))) -> c(x1) c(c(x1)) -> a(b(c(a(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: a(x1) -> b(x1) b(b(a(x1))) -> c(x1) c(c(x1)) -> a(c(b(a(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: A(x1) -> B(x1) B(b(a(x1))) -> C(x1) C(c(x1)) -> A(c(b(a(x1)))) C(c(x1)) -> C(b(a(x1))) C(c(x1)) -> B(a(x1)) C(c(x1)) -> A(x1) The TRS R consists of the following rules: a(x1) -> b(x1) b(b(a(x1))) -> c(x1) c(c(x1)) -> a(c(b(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) 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: C(c(x1)) -> C(b(a(x1))) C(c(x1)) -> B(a(x1)) C(c(x1)) -> A(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(B(x_1)) = x_1 POL(C(x_1)) = 2 + x_1 POL(a(x_1)) = 1 + x_1
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