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Proof step 70 from proof byte-mult-256 (previous step)

i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, b2i(i2b(256 * (i0 + i) + j0)) = b2i(i2b(j0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0)

Rule argument

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ i * j = j * i;% ⊦ i + j = j + i;