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Proof step 5 from proof gcd-mult (previous step)

∀ 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 * gcd(m, n), n0 * n ≤ n0 * m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m < n, ∀ 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 * gcd(m, n), n0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)% ⊦ m ≤ n ↔ ¬ n < m; (ss)