details

property	value
status	timeout (wallclock)
benchmark	40093.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	301.009315968 seconds
cpu usage	706.641767499
max memory	1.56123136E10

stage attributes

unavailable

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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 20 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 1347 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1))))))))))))))))))) 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(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: 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(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: 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(0(1(1(5(4(3(2(1(x1))))))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(1(5(4(3(2(1(x1))))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(5(4(3(2(1(x1)))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(x1) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))) The TRS R consists of the following rules: 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) Q is empty. 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: 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(5(4(3(2(1(x1)))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(1(5(4(3(2(1(x1))))))) 1^1(5(4(3(2(1(0(x1))))))) -> 1^1(x1) The TRS R consists of the following rules: 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) 1(5(4(3(2(1(0(x1))))))) -> 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) Q is empty. popout

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