details

property	value
status	complete
benchmark	tpa02.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	Secret_06_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.66861 seconds
cpu usage	6.55941
user time	6.26585
system time	0.29356
max virtual memory	1.8477524E7
max residence set size	479804.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 38 ms] (4) QDP (5) QDPBoundsTAProof [EQUIVALENT, 123 ms] (6) QDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(y, f(x, y)) -> f(f(a, y), f(a, x)) 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: F(y, f(x, y)) -> F(f(a, y), f(a, x)) F(y, f(x, y)) -> F(a, y) F(y, f(x, y)) -> F(a, x) The TRS R consists of the following rules: f(y, f(x, y)) -> f(f(a, y), f(a, x)) 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. F(y, f(x, y)) -> F(a, y) F(y, f(x, y)) -> F(a, x) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(F(x_1, x_2)) = [[0A]] + [[0A]] * x_1 + [[2A]] * x_2 >>> <<< POL(f(x_1, x_2)) = [[1A]] + [[1A]] * x_1 + [[1A]] * x_2 >>> <<< POL(a) = [[0A]] >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: f(y, f(x, y)) -> f(f(a, y), f(a, x)) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(y, f(x, y)) -> F(f(a, y), f(a, x)) The TRS R consists of the following rules: f(y, f(x, y)) -> f(f(a, y), f(a, x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPBoundsTAProof (EQUIVALENT) The DP-Problem (P, R) could be shown to be Match-(raise-)DP-Bounded [TAB_NONLEFTLINEAR] by 1 for the Rule: popout

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all output return to TRS Standard