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 Order [17,21] with Interpretation:

POL( 1 ) = max{0, -1}


POL( 0 ) = max{0, -1}


POL( s1(x1) ) = x1 + 1


POL( F3(x1, ..., x3) ) = x3 + 1



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.