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Proof step 3 from proof gcd-zero-01-h (previous step)

gcd(m, n) = 0 ∧ m < n, ∀ n0, m. n0 < n → gcd(m, n0) = 0 ∧ m < n0 → n0 = 0 ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
gcd(m, n) = 0, m < n, ∀ n0, m. n0 < n → gcd(m, n0) = 0 ∧ m < n0 → n0 = 0 ⊦ n = 0

Rule argument

Additional information