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	n034.star.cs.uiowa.edu
space	Gebhardt_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.59628200531 seconds
cpu usage	15.237839727
max memory	1.809907712E9

stage attributes

key	value
output-size	3825
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 9 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 50 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 186 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (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) 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: 0^1(0(0(0(x1)))) -> 0^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(1(0(1(x1)))) -> 0^1(0(1(0(x1)))) 0^1(1(0(1(x1)))) -> 0^1(1(0(x1))) 0^1(1(0(1(x1)))) -> 0^1(x1) 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. 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. 0^1(0(0(0(x1)))) -> 0^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(1(0(1(x1)))) -> 0^1(1(0(x1))) 0^1(1(0(1(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(0^1(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 0(1(0(1(x1)))) -> 0(0(1(0(x1)))) 0(0(0(0(x1)))) -> 1(0(0(1(x1)))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(0(1(x1)))) -> 0^1(0(1(0(x1)))) 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. popout

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