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TRS Standard pair #487068363
details
property
value
status
complete
benchmark
n007.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.79365 seconds
cpu usage
4.09557
user time
3.90701
system time
0.188552
max virtual memory
1.8345196E7
max residence set size
298024.0
stage attributes
key
value
starexec-result
NO
output
NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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) NonTerminationLoopProof [COMPLETE, 0 ms] (10) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y) -> f(x, x) f(s(x), 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(x, y) -> f(x, x) f(s(x), 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(x, y) -> f(x, x) f(s(x), y) -> f(y, x) The set Q consists of the following terms: f(x0, 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(x, y) -> F(x, x) F(s(x), y) -> F(y, x) The TRS R consists of the following rules: f(x, y) -> f(x, x) f(s(x), y) -> f(y, x) The set Q consists of the following terms: f(x0, 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: The TRS P consists of the following rules: F(x, y) -> F(x, x) F(s(x), y) -> F(y, x) R is empty. The set Q consists of the following terms: f(x0, x1)
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