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 strictly oriented and are deleted.


F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F1(x3)
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


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 strictly oriented and are deleted.


F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F2(x1, x2)
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > F2


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 strictly oriented 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 by weakly be oriented.

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F1(x2)
0  =  0
s1(x1)  =  s1(x1)
f3(x1, x2, x3)  =  f2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[0, f2] > [F1, s1]


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 strictly oriented and are deleted.


F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F1(x3)
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
F1 > s1
0 > s1


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 strictly oriented and are deleted.


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(x, y, f3(s1(x), s1(y), z))
The remaining pairs can at least by weakly be oriented.

F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
0  =  0
f3(x1, x2, x3)  =  f2(x1, x3)

Lexicographic Path Order [19].
Precedence:
0 > [s1, f2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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 strictly oriented and are deleted.


F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3)  =  F1(x3)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > F1


The following usable rules [14] were oriented: none



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