YES
proof of SK90_4.11.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 0 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


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

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   +(0, s(y)) -> s(y)
   s(+(0, y)) -> s(y)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:+_2 > s_1 > 0

and weight map:

   0=1
   s_1=1
   +_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   +(0, s(y)) -> s(y)
   s(+(0, y)) -> s(y)




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(4)
YES