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TRS Standard pair #487068243
details
property
value
status
complete
benchmark
2.05.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n179.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.45886 seconds
cpu usage
6.29222
user time
6.02309
system time
0.269136
max virtual memory
1.8477752E7
max residence set size
378792.0
stage attributes
key
value
starexec-result
NO
output
NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 68 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 0 ms] (8) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: +(x, +(y, z)) -> +(+(x, y), z) *(x, +(y, z)) -> +(*(x, y), *(x, z)) +(+(x, *(y, z)), *(y, u)) -> +(x, *(y, +(z, u))) 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: +^1(x, +(y, z)) -> +^1(+(x, y), z) +^1(x, +(y, z)) -> +^1(x, y) *^1(x, +(y, z)) -> +^1(*(x, y), *(x, z)) *^1(x, +(y, z)) -> *^1(x, y) *^1(x, +(y, z)) -> *^1(x, z) +^1(+(x, *(y, z)), *(y, u)) -> +^1(x, *(y, +(z, u))) +^1(+(x, *(y, z)), *(y, u)) -> *^1(y, +(z, u)) +^1(+(x, *(y, z)), *(y, u)) -> +^1(z, u) The TRS R consists of the following rules: +(x, +(y, z)) -> +(+(x, y), z) *(x, +(y, z)) -> +(*(x, y), *(x, z)) +(+(x, *(y, z)), *(y, u)) -> +(x, *(y, +(z, u))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. +^1(x, +(y, z)) -> +^1(x, y) *^1(x, +(y, z)) -> *^1(x, y) *^1(x, +(y, z)) -> *^1(x, z) +^1(+(x, *(y, z)), *(y, u)) -> *^1(y, +(z, u)) +^1(+(x, *(y, z)), *(y, u)) -> +^1(z, u) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( *^1_2(x_1, x_2) ) = 2x_2 POL( +^1_2(x_1, x_2) ) = x_1 + x_2 POL( +_2(x_1, x_2) ) = x_1 + x_2 + 1 POL( *_2(x_1, x_2) ) = 2x_2 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: +(+(x, *(y, z)), *(y, u)) -> +(x, *(y, +(z, u))) +(x, +(y, z)) -> +(+(x, y), z) *(x, +(y, z)) -> +(*(x, y), *(x, z)) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: +^1(x, +(y, z)) -> +^1(+(x, y), z) *^1(x, +(y, z)) -> +^1(*(x, y), *(x, z)) +^1(+(x, *(y, z)), *(y, u)) -> +^1(x, *(y, +(z, u))) The TRS R consists of the following rules:
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