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SRS Stand 10685 pair #381718086
details
property
value
status
complete
benchmark
16.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n024.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
6.57037091255 seconds
cpu usage
22.963070898
max memory
3.04943104E9
stage attributes
key
value
output-size
4632
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 11 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 81 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 75 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(0(1(x1)))) 0(1(0(1(x1)))) -> 0(0(1(0(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) Q is empty. ---------------------------------------- (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: 0^1(0(0(0(x1)))) -> 1^1(0(0(1(x1)))) 0^1(0(0(0(x1)))) -> 0^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(1(0(x1)))) -> 0^1(1(0(0(x1)))) 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 1^1(0(1(0(x1)))) -> 0^1(0(x1)) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(0(0(0(x1)))) -> 0^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 1^1(0(1(0(x1)))) -> 0^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( 0^1_1(x_1) ) = max{0, 2x_1 - 2} POL( 1^1_1(x_1) ) = max{0, 2x_1 - 2} POL( 1_1(x_1) ) = 2x_1 + 1 POL( 0_1(x_1) ) = 2x_1 + 1
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