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Proof step 6 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) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i, j ≤ i, 0 < i, 0 < j, ∀ 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

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs