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

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)
ACK(s(x), y) → F(x, x)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)
ACK(s(x), y) → F(x, x)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x0, x1)
F(x0, x1, x2)  =  F(x0, x1, x2)

Tags:
ACK has argument tags [3,4,3] and root tag 1
F has argument tags [0,4,3] and root tag 0

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACK(x₁, x₂)) = 0   
POL(F(x₁, x₂)) = x₂   
POL(ack(x₁, x₂)) = x₂   
POL(f(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), s(y)) → ACK(s(x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x0, x1)

Tags:
ACK has argument tags [0,1,0] and root tag 0

Comparison: DMS
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACK(x₁, x₂)) = x₂   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE