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Proof step 19 from proof gcd-div-gcd-pos (previous step)

i = (i0 * i3) * gcd(i, j), j = i2 * gcd(j, i), 0 ≤ i, 1 < i0, 0 < i0, 0 < j, 0 < i, ∃ i1. i0 * i1 = i2, 0 < i2, 0 < gcd(j, i), 0 ≤ j, 0 ≤ i2, 0 < i3, 0 ≤ i3, 0 < gcd(i, j) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult