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Proof step 35 from proof gcd-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1