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Proof step 41 from proof byte-mult-256 (previous step)

256 * (i0 + i) + j0 = (i0 + i - 1) * 256 + 256 + j0, i0 < 0, 0 < i, ~256 < j0, i0 < i, j0 < 0, i0 * 256 + j0 < 0, 0 < i0 + i ⊦ i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0), i0 + i = 0

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

Additional information