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Proof step 24 from proof gcd-div-gcd-pos (previous step)

i ≠ 0 ∨ j ≠ 0, j = (i0 * i1) * gcd(j, i), i = (i0 * i3) * gcd(i, j), 0 ≤ j, 0 < i, 0 < j, 0 < i0, 1 < i0, 0 ≤ i, 0 < gcd(i, j), 0 ≤ i3, 0 < i3, 0 ≤ i1, 0 < i1, 0 < gcd(j, i) ⊦ i0 * gcd(i, j) | i ∧ i0 * gcd(i, j) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; from specification int-mult%eq-same-02: ⊦ 0 < i → (i * j = i ↔ j = 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs