TRS Standard pair #516960795

details

property	value
status	complete
benchmark	wiehe09.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n172.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.24191999435 seconds
cpu usage	9.375726858
max memory	6.028288E8

stage attributes

key	value
output-size	15411
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 80 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 51 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 22 ms]
        (22) QDP
        (23) PisEmptyProof [EQUIVALENT, 0 ms]
        (24) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
   plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x))
   plus(zero, y) -> y
   plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y))))
   id(x) -> x
   if(true, x, y) -> x
   if(false, x, y) -> y
   not(x) -> if(x, false, true)
   gt(s(x), zero) -> true
   gt(zero, y) -> false
   gt(s(x), s(y)) -> gt(x, y)

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
   QUOT(s(x), s(y)) -> MINUS(x, y)
   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(minus(x, y), z) -> PLUS(y, z)
   PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
   PLUS(s(x), s(y)) -> IF(gt(x, y), x, y)
   PLUS(s(x), s(y)) -> GT(x, y)
   PLUS(s(x), s(y)) -> IF(not(gt(x, y)), id(x), id(y))
   PLUS(s(x), s(y)) -> NOT(gt(x, y))
   PLUS(s(x), s(y)) -> ID(x)
   PLUS(s(x), s(y)) -> ID(y)
   PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x))
   PLUS(s(x), x) -> IF(gt(x, x), id(x), id(x))
   PLUS(s(x), x) -> GT(x, x)
   PLUS(s(x), x) -> ID(x)
   PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y)))
   PLUS(id(x), s(y)) -> IF(gt(s(y), y), y, s(y))
   PLUS(id(x), s(y)) -> GT(s(y), y)
   NOT(x) -> IF(x, false, true)
   GT(s(x), s(y)) -> GT(x, y)

The TRS R consists of the following rules:

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job log

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