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TRS Standard pair #516966455
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
n004.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.97539591789 seconds
cpu usage
8.365518348
max memory
5.64973568E8
stage attributes
key
value
output-size
5030
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 83 ms]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:
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