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Proof step 3 from proof gcdhex (previous step)

j ≤ i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
j = 0, j ≤ i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ ∃ i0, j0. gcd(i, j) = i * i0 + j * j0
Proof step 6:
j ≤ i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Rule argument

fma: j = 0

Additional information

Used formulas: j = 0