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TRS Stand 20472 pair #381712677
details
property
value
status
complete
benchmark
test1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n077.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.88918185234 seconds
cpu usage
4.411814874
max memory
2.58555904E8
stage attributes
key
value
output-size
4400
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 42 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x), y) -> f(x, s(s(x))) f(x, s(s(y))) -> f(y, x) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is f(s(x), y) -> f(x, s(s(x))) f(x, s(s(y))) -> f(y, x) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x), y) -> f(x, s(s(x))) f(x, s(s(y))) -> f(y, x) The set Q consists of the following terms: f(s(x0), x1) f(x0, s(s(x1))) ---------------------------------------- (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(s(x), y) -> F(x, s(s(x))) F(x, s(s(y))) -> F(y, x) The TRS R consists of the following rules: f(s(x), y) -> f(x, s(s(x))) f(x, s(s(y))) -> f(y, x) The set Q consists of the following terms: f(s(x0), x1) f(x0, s(s(x1))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem:
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