details

property	value
status	complete
benchmark	17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n096.star.cs.uiowa.edu
space	Gebhardt_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.98375606537 seconds
cpu usage	15.629384755
max memory	1.348513792E9

stage attributes

key	value
output-size	4431
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) QDP (5) MRRProof [EQUIVALENT, 79 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 328 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(1(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(1(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(1(0(1(x1)))) 0^1(0(0(0(x1)))) -> 1^1(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(1(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) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: 0^1(0(0(0(x1)))) -> 1^1(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)) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 2 + x_1 POL(0^1(x_1)) = 3*x_1 POL(1(x_1)) = 2 + x_1 POL(1^1(x_1)) = 3*x_1 popout

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