details

property	value
status	complete
benchmark	z026.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n089.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	9.0945417881 seconds
cpu usage	32.481300158
max memory	1.629016064E9

stage attributes

key	value
output-size	2236
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 39 ms] (4) QDP (5) PisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(x1))) -> b(a(a(x1))) a(a(b(x1))) -> b(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(b(b(x1))) -> A(a(x1)) A(b(b(x1))) -> A(x1) A(a(b(x1))) -> A(x1) The TRS R consists of the following rules: a(b(b(x1))) -> b(a(a(x1))) a(a(b(x1))) -> b(b(a(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A(b(b(x1))) -> A(a(x1)) A(b(b(x1))) -> A(x1) A(a(b(x1))) -> A(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 ---------------------------------------- (4) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: a(b(b(x1))) -> b(a(a(x1))) a(a(b(x1))) -> b(b(a(x1))) 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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all output return to SRS Standard