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Proof step 48 from proof gcd-rec (previous step)

∃ i0. gcd(abs(i), abs(j)) * i0 = abs(j), gcd(abs(i), abs(j)) | abs(i), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0, gcd(abs(i), abs(j)) | j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
abs(i) = i0 * abs(j) + j0, gcd(abs(i), abs(j)) * i1 = abs(j), j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) | abs(i) ⊦ j = 0, gcd(abs(i), abs(j)) | j0

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1