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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
0  =  0
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• ACK(s(x), s(y)) → ACK(s(x), y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• F(x, y) → ACK(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 >= 2

• ACK(s(x), s(y)) → ACK(x, ack(s(x), y)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• ACK(s(x), y) → F(x, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 1 > 2

• ACK(s(x), 0) → ACK(x, s(0)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1

• F(s(x), y) → F(x, s(x)) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 1 >= 2

• F(x, s(y)) → F(y, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 1, 1 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(4) TRUE