YES
proof of SK90_2.15.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 63 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) -> 1
   f(s(x)) -> g(f(x))
   g(x) -> +(x, s(x))
   f(s(x)) -> +(f(x), s(f(x)))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
[f_1, 0, 1] > g_1 > [s_1, +_2]


Status:
f_1: multiset status
0: multiset status
1: multiset status
s_1: multiset status
g_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) -> 1
   f(s(x)) -> g(f(x))
   g(x) -> +(x, s(x))
   f(s(x)) -> +(f(x), s(f(x)))




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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