YES
proof of SK90_2.07.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 61 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:

   f(0, y) -> y
   f(x, 0) -> x
   f(i(x), y) -> i(x)
   f(f(x, y), z) -> f(x, f(y, z))
   f(g(x, y), z) -> g(f(x, z), f(y, z))
   f(1, g(x, y)) -> x
   f(2, g(x, y)) -> y

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
f_2 > [i_1, g_2]
0 > [i_1, g_2]
1 > [i_1, g_2]
2 > [i_1, g_2]


Status:
f_2: [1,2]
0: multiset status
i_1: multiset status
g_2: multiset status
1: multiset status
2: multiset status

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

   f(0, y) -> y
   f(x, 0) -> x
   f(i(x), y) -> i(x)
   f(f(x, y), z) -> f(x, f(y, z))
   f(g(x, y), z) -> g(f(x, z), f(y, z))
   f(1, g(x, y)) -> x
   f(2, g(x, y)) -> 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