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Proof step 2 from proof gcdhex (previous step)

0 ≤ i ∧ 0 ≤ j ∧ j ≤ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs