details

property	value
status	complete
benchmark	bn111.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n173.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	14.6949150562 seconds
cpu usage	36.885824903
max memory	3.296088064E9

stage attributes

key	value
output-size	2453
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 36 ms] (4) QDP (5) PisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(X, Y), Z) -> f(X, f(Y, Z)) f(X, f(Y, Z)) -> f(Y, Y) 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(f(X, Y), Z) -> F(X, f(Y, Z)) F(f(X, Y), Z) -> F(Y, Z) F(X, f(Y, Z)) -> F(Y, Y) The TRS R consists of the following rules: f(f(X, Y), Z) -> f(X, f(Y, Z)) f(X, f(Y, Z)) -> f(Y, Y) 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(f(X, Y), Z) -> F(X, f(Y, Z)) F(f(X, Y), Z) -> F(Y, Z) F(X, f(Y, Z)) -> F(Y, Y) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(F(x_1, x_2)) = x_1*x_2 + 2*x_1^2 + 2*x_2 POL(f(x_1, x_2)) = 2 + 2*x_1 + x_1*x_2 + 2*x_1^2 + 2*x_2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: f(X, f(Y, Z)) -> f(Y, Y) f(f(X, Y), Z) -> f(X, f(Y, Z)) ---------------------------------------- (4) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: f(f(X, Y), Z) -> f(X, f(Y, Z)) f(X, f(Y, Z)) -> f(Y, Y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (6) YES popout

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