YES
proof of SK90_2.01.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 69 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

   i(0) -> 0
   +(0, y) -> y
   +(x, 0) -> x
   i(i(x)) -> x
   +(i(x), x) -> 0
   +(x, i(x)) -> 0
   i(+(x, y)) -> +(i(x), i(y))
   +(x, +(y, z)) -> +(+(x, y), z)
   +(+(x, i(y)), y) -> x
   +(+(x, y), i(y)) -> x

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(+(x_1, x_2)) = 2 + x_1 + 2*x_2
   POL(0) = 2
   POL(i(x_1)) = 2*x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   i(0) -> 0
   +(0, y) -> y
   +(x, 0) -> x
   i(+(x, y)) -> +(i(x), i(y))
   +(x, +(y, z)) -> +(+(x, y), z)
   +(+(x, i(y)), y) -> x
   +(+(x, y), i(y)) -> x




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   i(i(x)) -> x
   +(i(x), x) -> 0
   +(x, i(x)) -> 0

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:+_2 > 0 > i_1

and weight map:

   0=3
   i_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:

   i(i(x)) -> x
   +(i(x), x) -> 0
   +(x, i(x)) -> 0




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES