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Proof step 10 from proof gcd-zero-01-h (previous step)

gcd(m, n1) = 0, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → n0 = 0 ⊦ gcd(m, n1) = 0 ∧ m < n1, n1 + m = 0, n1 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 11:
gcd(m, n1) = 0, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → n0 = 0 ⊦ m < n1, n1 + m = 0, n1 = 0

Rule argument

pos: right 1

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