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Proof step 3 from proof div-mult-neg (previous step)

j * i + i0 = ~(j * ~ i + ~ i0), abs(i0) < j, 0 < j, i ≤ 0, i < j, i0 ≤ 0, i0 < j ⊦ (j * i + i0) / j = i

User applied rule insert equation

And obtained the following new goals

Proof step 4:
abs(i0) < j, 0 < j, i ≤ 0, i < j, i0 ≤ 0, i0 < j ⊦ ~(j * ~ i + ~ i0) / j = i

Rule argument

pos: left 1 flags: discard

Additional information