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(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, 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(x, y, s1(z)) -> F3(0, 1, z)
F3(0, 1, x) -> F3(s1(x), x, x)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))

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:

F3(x, y, s1(z)) -> F3(0, 1, z)
F3(0, 1, x) -> F3(s1(x), x, x)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, 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(x, y, s1(z)) -> F3(0, 1, z)
The remaining pairs can at least be oriented weakly.

F3(0, 1, x) -> F3(s1(x), x, x)
Used ordering: Polynomial interpretation [21]:

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

F3(0, 1, x) -> F3(s1(x), x, x)

The TRS R consists of the following rules:

f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.