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TRS Standard pair #487068748
details
property
value
status
complete
benchmark
z14.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n174.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.81521 seconds
cpu usage
3.99067
user time
3.82656
system time
0.164104
max virtual memory
1.8409712E7
max residence set size
242284.0
stage attributes
key
value
starexec-result
YES
output
YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0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) QDPToSRSProof [SOUND, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) MRRProof [EQUIVALENT, 0 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(a(b(x))), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) 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, a(b(y))) -> f(a(a(b(x))), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(a(b(x))), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The set Q consists of the following terms: f(x0, a(b(x1))) f(a(x0), x1) f(b(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, a(b(y))) -> F(a(a(b(x))), y) F(a(x), y) -> F(x, a(y)) F(b(x), y) -> F(x, b(y)) The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(a(b(x))), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The set Q consists of the following terms: f(x0, a(b(x1))) f(a(x0), x1) f(b(x0), x1) We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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