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forall m in N, n in N: (m != 0 \/ n != 0) => Euclid(m, n) = gcd(m, n).Proof by complete induction on term m+n.
We take arbitrary m in N and n in N and assume
We have to prove
(1) forall x in N, y in N: x+y < m+n => (x != 0 \/ y != 0) => Euclid(x, y) = gcd(x, y).
We assume (3) (m != 0 \/ n != 0) and prove (4) Euclid(m, n) = gcd(m, n).
(2) (m != 0 \/ n != 0) => Euclid(m, n) = gcd(m, n).