details

property	value
status	complete
benchmark	02.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.9154 seconds
cpu usage	19.9151
user time	18.9867
system time	0.92839
max virtual memory	5.9601472E7
max residence set size	2359892.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 6 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 89 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 85 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(a(a(b(x1)))) b(a(b(x1))) -> a(b(a(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(a(a(a(x1)))) -> B(a(a(b(x1)))) A(a(a(a(x1)))) -> A(a(b(x1))) A(a(a(a(x1)))) -> A(b(x1)) A(a(a(a(x1)))) -> B(x1) B(a(b(x1))) -> A(b(a(x1))) B(a(b(x1))) -> B(a(x1)) B(a(b(x1))) -> A(x1) The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(a(a(b(x1)))) b(a(b(x1))) -> a(b(a(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(a(a(a(x1)))) -> A(a(b(x1))) A(a(a(a(x1)))) -> A(b(x1)) A(a(a(a(x1)))) -> B(x1) B(a(b(x1))) -> B(a(x1)) B(a(b(x1))) -> A(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A_1(x_1) ) = max{0, x_1 - 1} POL( B_1(x_1) ) = max{0, x_1 - 1} POL( b_1(x_1) ) = 2x_1 + 1 POL( a_1(x_1) ) = 2x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(a(b(x1))) -> a(b(a(x1))) a(a(a(a(x1)))) -> b(a(a(b(x1)))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(a(x1)))) -> B(a(a(b(x1)))) B(a(b(x1))) -> A(b(a(x1))) The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(a(a(b(x1)))) b(a(b(x1))) -> a(b(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. popout

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