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:

g2(x, y) -> x
g2(x, y) -> y
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:

g2(x, y) -> x
g2(x, y) -> y
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))

Q is empty.

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:

g2(x, y) -> x
g2(x, y) -> y
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
      ↳ QDPAfsSolverProof

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:

g2(x, y) -> x
g2(x, y) -> y
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(x, y, s1(z)) -> F3(0, 1, z)
Used argument filtering: F3(x1, x2, x3)  =  x3
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
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:

g2(x, y) -> x
g2(x, y) -> y
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 contains 0 SCCs with 1 less node.