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Proof step 2 from proof gcdexh (previous step)

n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
n = 0, n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0
Proof step 5:
n ≤ m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ n = 0, ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

Rule argument

fma: n = 0

Additional information

Used formulas: n = 0