details

property	value
status	complete
benchmark	jwmatchb2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Waldmann_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	13.825 seconds
cpu usage	29.1497
user time	27.6122
system time	1.53749
max virtual memory	1.8703764E7
max residence set size	4218912.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 597 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(f(x, y)) -> f(f(a, h(h(y))), 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: h(f(x, y)) -> f(f(a, h(h(y))), x) The set Q consists of the following terms: h(f(x0, x1)) ---------------------------------------- (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: H(f(x, y)) -> H(h(y)) H(f(x, y)) -> H(y) The TRS R consists of the following rules: h(f(x, y)) -> f(f(a, h(h(y))), x) The set Q consists of the following terms: h(f(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule H(f(x, y)) -> H(h(y)) at position [0] we obtained the following new rules [LPAR04]: (H(f(y0, f(x0, x1))) -> H(f(f(a, h(h(x1))), x0)),H(f(y0, f(x0, x1))) -> H(f(f(a, h(h(x1))), x0))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: H(f(x, y)) -> H(y) H(f(y0, f(x0, x1))) -> H(f(f(a, h(h(x1))), x0)) The TRS R consists of the following rules: h(f(x, y)) -> f(f(a, h(h(y))), x) The set Q consists of the following terms: h(f(x0, x1)) popout

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all output return to TRS Standard