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:

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

The set Q consists of the following terms:

gcd(x0, 0)
gcd(0, x0)
gcd(s(x0), s(x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

GCD(s(x), s(y)) → GCD(-(x, y), s(y))
GCD(s(x), s(y)) → GCD(s(x), -(y, x))

The TRS R consists of the following rules:

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

The set Q consists of the following terms:

gcd(x0, 0)
gcd(0, x0)
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

GCD(s(x), s(y)) → GCD(-(x, y), s(y))
GCD(s(x), s(y)) → GCD(s(x), -(y, x))

The TRS R consists of the following rules:

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

The set Q consists of the following terms:

gcd(x0, 0)
gcd(0, x0)
gcd(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

GCD(s(x), s(y)) → GCD(-(x, y), s(y))
GCD(s(x), s(y)) → GCD(s(x), -(y, x))

The TRS R consists of the following rules:

gcd(x, 0) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if(<(x, y), gcd(s(x), -(y, x)), gcd(-(x, y), s(y)))

The set Q consists of the following terms:

gcd(x0, 0)
gcd(0, x0)
gcd(s(x0), s(x1))

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.