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Proof step 4 from proof gcdexh (previous step)

∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0. m * m0 + m = 0 ∨ m = 0 ∨ m0 = 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 1 abort: F discard: T computed: F

Additional information

Used terms: 1