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Proof step 102 from proof byte-mult-256 (previous step)

~256 < i1 ∧ i1 < ~128, i1 < 0, ~256 ≤ i1, 0 ≤ i0, i < 0, i0 + i < ~1, i0 + i < 0 ⊦ i0 + i = 0, i = 0, i1 + 256 = b2i(i2b(i1 + 256))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 103:
i0 + i < 0, i0 + i < ~1, i < 0, 0 ≤ i0, ~256 ≤ i1, i1 < 0, i1 < ~128, ~256 < i1 ⊦ i1 + 256 = b2i(i2b(i1 + 256)), i = 0, i0 + i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;