Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
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(s(x), y)
F(x, s(y)) → F(y, x)
F(s(x), y) → F(x, s(x))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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(s(x), y)
F(x, s(y)) → F(y, x)
F(s(x), y) → F(x, s(x))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACK(s(x), 0) → ACK(x, s(0))
F(x, s(y)) → F(y, x)
F(s(x), y) → F(x, s(x))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), y) → F(x, x)
The remaining pairs can at least be oriented weakly.

ACK(s(x), s(y)) → ACK(s(x), y)
F(x, y) → ACK(x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = 3/2 + (3/4)x_1 + (1/4)x_2   
POL(ACK(x1, x2)) = 4 + (3/4)x_1   
POL(s(x1)) = 1/4 + (4)x_1   
POL(F(x1, x2)) = 4 + (3/4)x_1 + (1/2)x_2   
POL(0) = 0   
POL(ack(x1, x2)) = 7/2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

ACK(s(x), s(y)) → ACK(s(x), y)
F(x, y) → ACK(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ACK(x1, x2)) = (4)x_2   
POL(s(x1)) = 1 + (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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