details

property	value
status	complete
benchmark	size-11-alpha-2-num-2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	Waldmann_07_size11

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.69965195656 seconds
cpu usage	27.447128745
max memory	3.93455616E9

stage attributes

key	value
output-size	5026
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 23 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 144 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 75 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 44 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(b(b(a(x1)))) -> a(a(b(a(b(b(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: A(b(b(a(x1)))) -> A(a(b(a(b(b(x1)))))) A(b(b(a(x1)))) -> A(b(a(b(b(x1))))) A(b(b(a(x1)))) -> A(b(b(x1))) The TRS R consists of the following rules: a(x1) -> x1 a(b(b(a(x1)))) -> a(a(b(a(b(b(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. A(b(b(a(x1)))) -> A(b(b(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(A(x_1)) = [[0A]] + [[0A, -I, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[0A], [-I], [-I]] + [[-I, -I, 0A], [-I, -I, -I], [-I, 0A, -I]] * x_1 >>> <<< POL(a(x_1)) = [[0A], [1A], [0A]] + [[0A, 0A, -I], [1A, 1A, -I], [0A, 0A, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(x1) -> x1 a(b(b(a(x1)))) -> a(a(b(a(b(b(x1)))))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(b(a(x1)))) -> A(a(b(a(b(b(x1)))))) A(b(b(a(x1)))) -> A(b(a(b(b(x1))))) popout

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all output return to SRS Stand 10685