details

property	value
status	complete
benchmark	z01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n031.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.00928997993 seconds
cpu usage	8.462848996
max memory	6.214656E8

stage attributes

key	value
output-size	11914
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) ATransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) RFCMatchBoundsDPProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(f, a(f, a(g, a(g, x)))) -> a(g, a(g, a(g, a(f, a(f, a(f, x)))))) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(f, a(f, a(g, a(g, x)))) -> a(g, a(g, a(g, a(f, a(f, a(f, x)))))) The set Q consists of the following terms: a(f, a(f, a(g, a(g, x0)))) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(f, a(f, a(g, a(g, x)))) -> A(g, a(g, a(g, a(f, a(f, a(f, x)))))) A(f, a(f, a(g, a(g, x)))) -> A(g, a(g, a(f, a(f, a(f, x))))) A(f, a(f, a(g, a(g, x)))) -> A(g, a(f, a(f, a(f, x)))) A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, a(f, x))) A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x)) A(f, a(f, a(g, a(g, x)))) -> A(f, x) The TRS R consists of the following rules: a(f, a(f, a(g, a(g, x)))) -> a(g, a(g, a(g, a(f, a(f, a(f, x)))))) The set Q consists of the following terms: a(f, a(f, a(g, a(g, x0)))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(f, a(f, a(g, a(g, x)))) -> A(f, a(f, x)) popout

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