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Proof step 13 from proof gcd-div-gcd-pos (previous step)

abs(j0) < gcd(j, i), i = i1 * gcd(i, j), j = i2 * gcd(j, i) ∧ j0 = 0, 0 ≤ j, 0 ≤ i2, 0 ≤ j0, 0 < i1, 0 ≤ i, 0 < gcd(i, j), 0 ≤ i1, 1 < i0, 0 < i0, 0 < j, 0 < i, i0 | i2, i0 | i1, j0 < gcd(i, j) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
j = i2 * gcd(j, i), i = i1 * gcd(i, j), i0 | i1, i0 | i2, 0 < i, 0 < j, 0 < i0, 1 < i0, 0 ≤ i1, 0 < gcd(i, j), 0 ≤ i, 0 < i1, 0 ≤ i2, 0 ≤ j, 0 < gcd(j, i), 0 < i2 ⊦

Rule argument

Additional information

Used simplifier rules: 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%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