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
↳ 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)))
Q is empty.
Using Dependency Pairs [1,15] 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)))
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:
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)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.