details

property	value
status	complete
benchmark	kusakari1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Mixed_AC

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.76937 seconds
cpu usage	6.43716
user time	6.2028
system time	0.234352
max virtual memory	1.8308428E7
max residence set size	332276.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ETRS could be proven: (0) ETRS (1) EquationalDependencyPairsProof [EQUIVALENT, 17 ms] (2) EDP (3) ERuleRemovalProof [EQUIVALENT, 22 ms] (4) EDP (5) EDPPoloProof [EQUIVALENT, 15 ms] (6) EDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Equational rewrite system: The TRS R consists of the following rules: +(g(x), g(y)) -> g(+(g(a), +(x, y))) The set E consists of the following equations: +(x, y) == +(y, x) +(+(x, y), z) == +(x, +(y, z)) ---------------------------------------- (1) EquationalDependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,DA_STEIN] we result in the following initial EDP problem: The TRS P consists of the following rules: +^1(g(x), g(y)) -> +^1(g(a), +(x, y)) +^1(g(x), g(y)) -> +^1(x, y) +^1(+(g(x), g(y)), ext) -> +^1(g(+(g(a), +(x, y))), ext) +^1(+(g(x), g(y)), ext) -> +^1(g(a), +(x, y)) +^1(+(g(x), g(y)), ext) -> +^1(x, y) The TRS R consists of the following rules: +(g(x), g(y)) -> g(+(g(a), +(x, y))) +(+(g(x), g(y)), ext) -> +(g(+(g(a), +(x, y))), ext) The set E consists of the following equations: +(x, y) == +(y, x) +(+(x, y), z) == +(x, +(y, z)) The set E# consists of the following equations: +^1(x, y) == +^1(y, x) +^1(+(x, y), z) == +^1(x, +(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (2) Obligation: The TRS P consists of the following rules: +^1(g(x), g(y)) -> +^1(g(a), +(x, y)) +^1(g(x), g(y)) -> +^1(x, y) +^1(+(g(x), g(y)), ext) -> +^1(g(+(g(a), +(x, y))), ext) +^1(+(g(x), g(y)), ext) -> +^1(g(a), +(x, y)) +^1(+(g(x), g(y)), ext) -> +^1(x, y) The TRS R consists of the following rules: +(g(x), g(y)) -> g(+(g(a), +(x, y))) +(+(g(x), g(y)), ext) -> +(g(+(g(a), +(x, y))), ext) The set E consists of the following equations: +(x, y) == +(y, x) +(+(x, y), z) == +(x, +(y, z)) The set E# consists of the following equations: +^1(x, y) == +^1(y, x) +^1(+(x, y), z) == +^1(x, +(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (3) ERuleRemovalProof (EQUIVALENT) By using the rule removal processor [DA_STEIN] with the following polynomial ordering [POLO], at least one Dependency Pair or term rewrite system rule of this EDP problem can be strictly oriented. Strictly oriented dependency pairs: +^1(g(x), g(y)) -> +^1(g(a), +(x, y)) +^1(g(x), g(y)) -> +^1(x, y) +^1(+(g(x), g(y)), ext) -> +^1(g(a), +(x, y)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Equational