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Proof step 39 from proof gcd-rec (previous step)

abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), j0 < abs(j), 0 ≤ i0, 0 ≤ j0, gcd(abs(i), abs(j)) | abs(j) ∧ gcd(abs(i), abs(j)) | j0 ∧ 0 ≠ abs(j) → gcd(abs(i), abs(j)) ≤ gcd(abs(j), j0) ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
abs(i) = i0 * abs(j) + j0, j0 < abs(j), gcd(abs(j), j0) < gcd(abs(i), abs(j)), 0 ≤ j0, 0 ≤ i0 ⊦ j = 0, gcd(abs(i), abs(j)) | abs(j) ∧ gcd(abs(i), abs(j)) | j0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ abs(i) = 0 ↔ i = 0; (s)