details

property	value
status	complete
benchmark	4046.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n171.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.2367079258 seconds
cpu usage	9.784357686
max memory	1.053077504E9

stage attributes

key	value
output-size	8226
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 65 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 4(2(4(x1))) -> 2(0(0(5(3(3(5(2(0(4(x1)))))))))) 4(4(2(4(2(x1))))) -> 2(0(5(2(1(4(0(2(0(1(x1)))))))))) 0(5(4(2(4(3(x1)))))) -> 5(1(5(5(3(5(3(0(0(0(x1)))))))))) 1(1(4(5(3(3(x1)))))) -> 1(3(1(1(3(0(1(2(2(1(x1)))))))))) 3(1(4(3(1(2(x1)))))) -> 0(0(1(1(4(2(3(0(0(3(x1)))))))))) 3(2(4(2(4(1(x1)))))) -> 0(2(1(1(1(5(3(1(3(3(x1)))))))))) 3(3(0(4(1(2(x1)))))) -> 3(5(1(2(0(2(0(5(3(1(x1)))))))))) 4(1(4(5(0(5(4(x1))))))) -> 4(1(5(3(1(0(5(3(1(0(x1)))))))))) 4(4(0(5(4(2(2(x1))))))) -> 4(0(4(3(4(4(4(5(4(1(x1)))))))))) 5(4(5(3(2(4(3(x1))))))) -> 2(5(5(5(0(4(5(0(1(4(x1)))))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 4(2(4(x1))) -> 4(0(2(5(3(3(5(0(0(2(x1)))))))))) 2(4(2(4(4(x1))))) -> 1(0(2(0(4(1(2(5(0(2(x1)))))))))) 3(4(2(4(5(0(x1)))))) -> 0(0(0(3(5(3(5(5(1(5(x1)))))))))) 3(3(5(4(1(1(x1)))))) -> 1(2(2(1(0(3(1(1(3(1(x1)))))))))) 2(1(3(4(1(3(x1)))))) -> 3(0(0(3(2(4(1(1(0(0(x1)))))))))) 1(4(2(4(2(3(x1)))))) -> 3(3(1(3(5(1(1(1(2(0(x1)))))))))) 2(1(4(0(3(3(x1)))))) -> 1(3(5(0(2(0(2(1(5(3(x1)))))))))) 4(5(0(5(4(1(4(x1))))))) -> 0(1(3(5(0(1(3(5(1(4(x1)))))))))) 2(2(4(5(0(4(4(x1))))))) -> 1(4(5(4(4(4(3(4(0(4(x1)))))))))) 3(4(2(3(5(4(5(x1))))))) -> 4(1(0(5(4(0(5(5(5(2(x1)))))))))) Q is empty. ---------------------------------------- (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: 4^1(2(4(x1))) -> 4^1(0(2(5(3(3(5(0(0(2(x1)))))))))) 4^1(2(4(x1))) -> 2^1(5(3(3(5(0(0(2(x1)))))))) 4^1(2(4(x1))) -> 3^1(3(5(0(0(2(x1)))))) 4^1(2(4(x1))) -> 3^1(5(0(0(2(x1))))) 4^1(2(4(x1))) -> 2^1(x1) 2^1(4(2(4(4(x1))))) -> 1^1(0(2(0(4(1(2(5(0(2(x1)))))))))) 2^1(4(2(4(4(x1))))) -> 2^1(0(4(1(2(5(0(2(x1)))))))) 2^1(4(2(4(4(x1))))) -> 4^1(1(2(5(0(2(x1)))))) 2^1(4(2(4(4(x1))))) -> 1^1(2(5(0(2(x1))))) 2^1(4(2(4(4(x1))))) -> 2^1(5(0(2(x1)))) 2^1(4(2(4(4(x1))))) -> 2^1(x1) 3^1(4(2(4(5(0(x1)))))) -> 3^1(5(3(5(5(1(5(x1))))))) 3^1(4(2(4(5(0(x1)))))) -> 3^1(5(5(1(5(x1))))) 3^1(4(2(4(5(0(x1)))))) -> 1^1(5(x1)) 3^1(3(5(4(1(1(x1)))))) -> 1^1(2(2(1(0(3(1(1(3(1(x1)))))))))) 3^1(3(5(4(1(1(x1)))))) -> 2^1(2(1(0(3(1(1(3(1(x1))))))))) 3^1(3(5(4(1(1(x1)))))) -> 2^1(1(0(3(1(1(3(1(x1)))))))) 3^1(3(5(4(1(1(x1)))))) -> 1^1(0(3(1(1(3(1(x1))))))) 3^1(3(5(4(1(1(x1)))))) -> 3^1(1(1(3(1(x1))))) popout

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all output return to SRS Standard