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:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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:

F3(s1(0), y, z) -> F3(0, s1(y), s1(z))
F3(s1(x), 0, s1(z)) -> F3(x, s1(0), z)
F3(s1(x), s1(y), 0) -> F3(x, y, s1(0))
F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, y, f3(0, s1(s1(y)), s1(z)))
F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)
F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))
F3(s1(x), s1(y), s1(z)) -> F3(x, y, f3(s1(x), s1(y), z))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F3(s1(0), y, z) -> F3(0, s1(y), s1(z))
F3(s1(x), 0, s1(z)) -> F3(x, s1(0), z)
F3(s1(x), s1(y), 0) -> F3(x, y, s1(0))
F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, y, f3(0, s1(s1(y)), s1(z)))
F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)
F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))
F3(s1(x), s1(y), s1(z)) -> F3(x, y, f3(s1(x), s1(y), z))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 1 less node.

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

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

F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F3(x1, x2, x3)) = 3·x3   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.

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

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

F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F3(x1, x2, x3)) = 3·x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.

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

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

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, y, f3(0, s1(s1(y)), s1(z)))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, y, f3(0, s1(s1(y)), s1(z)))
The remaining pairs can at least be oriented weakly.

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F3(x1, x2, x3)) = 3·x2   
POL(f3(x1, x2, x3)) = x2   
POL(s1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



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

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

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F3(x1, x2, x3)) = x3   
POL(s1(x1)) = 1 + 3·x1   

The following usable rules [14] were oriented: none



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

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

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.

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

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

F3(s1(x), s1(y), 0) -> F3(x, y, s1(0))
F3(s1(x), 0, s1(z)) -> F3(x, s1(0), z)
F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
F3(s1(x), s1(y), s1(z)) -> F3(x, y, f3(s1(x), s1(y), z))

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(s1(x), s1(y), 0) -> F3(x, y, s1(0))
F3(s1(x), s1(y), s1(z)) -> F3(x, y, f3(s1(x), s1(y), z))
The remaining pairs can at least be oriented weakly.

F3(s1(x), 0, s1(z)) -> F3(x, s1(0), z)
F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(F3(x1, x2, x3)) = 3·x1 + 3·x2   
POL(f3(x1, x2, x3)) = 1 + 3·x1 + 2·x3   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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

F3(s1(x), 0, s1(z)) -> F3(x, s1(0), z)
F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
                    ↳ QDPOrderProof

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

F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)

The TRS R consists of the following rules:

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.


F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F3(x1, x2, x3)) = 3·x3   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

f3(x, 0, 0) -> s1(x)
f3(0, y, 0) -> s1(y)
f3(0, 0, z) -> s1(z)
f3(s1(0), y, z) -> f3(0, s1(y), s1(z))
f3(s1(x), s1(y), 0) -> f3(x, y, s1(0))
f3(s1(x), 0, s1(z)) -> f3(x, s1(0), z)
f3(0, s1(0), s1(0)) -> s1(s1(0))
f3(s1(x), s1(y), s1(z)) -> f3(x, y, f3(s1(x), s1(y), z))
f3(0, s1(s1(y)), s1(0)) -> f3(0, y, s1(0))
f3(0, s1(0), s1(s1(z))) -> f3(0, s1(0), z)
f3(0, s1(s1(y)), s1(s1(z))) -> f3(0, y, f3(0, s1(s1(y)), s1(z)))

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.