YES
proof of /TPDB/AG01_#3.26.ari
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) MNOCProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 15 ms]
(6) QDP
(7) PisEmptyProof [EQUIVALENT, 3 ms]
(8) YES


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

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> F(f(x))
   F(s(s(x))) -> F(x)

The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(s(x))) -> F(f(x))
   F(s(s(x))) -> F(x)

The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

The set Q consists of the following terms:

   f(x0)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   F(s(s(x))) -> F(f(x))
   F(s(s(x))) -> F(x)


Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1)) = 2*x_1
   POL(f(x_1)) = 2 + x_1
   POL(s(x_1)) = 2 + x_1


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(6)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))

The set Q consists of the following terms:

   f(x0)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(8)
YES