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Proof step 17 from proof gcd-mult (previous step)

n0 * n ≤ n0 * m, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m - n0 * n, n0 * n) = n0 * gcd(m, n)

User applied rule cut formula

And obtained the following new goals

Proof step 18:
n0 * n ≤ n0 * m, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ n0 = 0, gcd(n0 * m - n0 * n, n0 * n) = n0 * gcd(m, n)

Rule argument

fma: n0 = 0

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, n) = n;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%mult-base: ⊦ m * 0 = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used formulas: n0 = 0