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Proof step 29 from proof gcd-rec (previous step)

gcd(abs(j), j0) * i1 = j0, abs(i) = i0 * abs(j) + j0, ∃ i0. gcd(abs(j), j0) * i0 = abs(j), j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(abs(i), abs(j)) < gcd(abs(j), j0) ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), j = 0, i = 0, gcd(abs(j), j0) | abs(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
gcd(abs(j), j0) * i2 = abs(j), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) * i1 = j0, gcd(abs(i), abs(j)) < gcd(abs(j), j0), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ i = 0, j = 0, gcd(abs(i), abs(j)) = gcd(abs(j), j0), gcd(abs(j), j0) | abs(i)

Rule argument

Additional information

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