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.

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 contains 0 SCCs with 2 less nodes.