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:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(x, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(x, x)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
F2(s1(x), y) -> F2(x, s1(x))
F2(x, s1(y)) -> F2(y, x)
ACK2(s1(x), y) -> F2(x, x)
F2(x, y) -> ACK2(x, y)
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(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:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
F2(s1(x), y) -> F2(x, s1(x))
F2(x, s1(y)) -> F2(y, x)
ACK2(s1(x), y) -> F2(x, x)
F2(x, y) -> ACK2(x, y)
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
F2(s1(x), y) -> F2(x, s1(x))
F2(x, s1(y)) -> F2(y, x)
ACK2(s1(x), y) -> F2(x, x)
ACK2(s1(x), 0) -> ACK2(x, s1(0))
The remaining pairs can at least be oriented weakly.

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
F2(x, y) -> ACK2(x, y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(ACK2(x1, x2)) = 1 + 3·x1   
POL(F2(x1, x2)) = 1 + 3·x1 + x2   
POL(ack2(x1, x2)) = 0   
POL(f2(x1, x2)) = 0   
POL(s1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented: none



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

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

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
F2(x, y) -> ACK2(x, y)

The TRS R consists of the following rules:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)

The TRS R consists of the following rules:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACK2(x1, x2)) = 3·x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] 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:

ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
f2(s1(x), y) -> f2(x, s1(x))
f2(x, s1(y)) -> f2(y, x)
f2(x, y) -> ack2(x, y)
ack2(s1(x), y) -> f2(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.