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.

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 contains 4 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(0, s1(0), s1(s1(z))) -> F3(0, s1(0), z)
Used argument filtering: F3(x1, x2, x3)  =  x3
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(0, s1(s1(y)), s1(0)) -> F3(0, y, s1(0))
Used argument filtering: F3(x1, x2, x3)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, y, f3(0, s1(s1(y)), s1(z)))
Used argument filtering: F3(x1, x2, x3)  =  x2
s1(x1)  =  s1(x1)
f3(x1, x2, x3)  =  f3(x1, x2, x3)
0  =  0
Used ordering: Quasi Precedence: f_3 > s_1


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(0, s1(s1(y)), s1(s1(z))) -> F3(0, s1(s1(y)), s1(z))
Used argument filtering: F3(x1, x2, x3)  =  x3
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))
Used argument filtering: F3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
f3(x1, x2, x3)  =  f3(x1, x2, x3)
0  =  0
Used ordering: Quasi Precedence: f_3 > s_1


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(s1(x), s1(y), s1(z)) -> F3(s1(x), s1(y), z)
Used argument filtering: F3(x1, x2, x3)  =  x3
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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.