Termination w.r.t. Q proof of TRS/SK904.18.trs

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:

gcd₂(x, 0) -> x
gcd₂(0, y) -> y
gcd₂(s₁(x), s₁(y)) -> if₃(<₂(x, y), gcd₂(s₁(x), -₂(y, x)), gcd₂(-₂(x, y), s₁(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

gcd₂(x, 0) -> x
gcd₂(0, y) -> y
gcd₂(s₁(x), s₁(y)) -> if₃(<₂(x, y), gcd₂(s₁(x), -₂(y, x)), gcd₂(-₂(x, y), s₁(y)))

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:

GCD₂(s₁(x), s₁(y)) -> GCD₂(-₂(x, y), s₁(y))
GCD₂(s₁(x), s₁(y)) -> GCD₂(s₁(x), -₂(y, x))

The TRS R consists of the following rules:

gcd₂(x, 0) -> x
gcd₂(0, y) -> y
gcd₂(s₁(x), s₁(y)) -> if₃(<₂(x, y), gcd₂(s₁(x), -₂(y, x)), gcd₂(-₂(x, y), s₁(y)))

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:

GCD₂(s₁(x), s₁(y)) -> GCD₂(-₂(x, y), s₁(y))
GCD₂(s₁(x), s₁(y)) -> GCD₂(s₁(x), -₂(y, x))

The TRS R consists of the following rules:

gcd₂(x, 0) -> x
gcd₂(0, y) -> y
gcd₂(s₁(x), s₁(y)) -> if₃(<₂(x, y), gcd₂(s₁(x), -₂(y, x)), gcd₂(-₂(x, y), s₁(y)))

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 2 less nodes.