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

gcd(abs(j), j0) * i1 = j0, gcd(abs(j), j0) * i2 = abs(j), abs(i) = i0 * abs(j) + j0, gcd(i0 * abs(j) + j0, abs(j)) < gcd(abs(j), j0), j0 < abs(j), 0 ≤ i0, 0 ≤ j0 ⊦ gcd(abs(j), j0) * (i2 * i0 + i1) = i0 * abs(j) + j0, gcd(i0 * abs(j) + j0, abs(j)) = gcd(abs(j), j0), i = 0, j = 0, ∃ i1. gcd(abs(j), j0) * i1 = i0 * abs(j) + j0

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: rotate, discard, [[2,1,2,2]]

Additional information

Used simplifier rules: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%neq-not-zero: ⊦ i * j ≠ i * j0 → i ≠ 0; (forward) from specification int-mult% ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); (s)