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Proof step 91 from proof byte-mult-256 (previous step)

i0 + i + 1 < 0 ∧ j0 - 256 < 0, i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b(j0 - 256) = i2b(j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 92:
i0 + i < ~1, i < 0, 0 ≤ i0, 0 ≤ j0, j0 < 256, i0 + i < 0 ⊦ i2b(j0 - 256) = i2b(j0), i = 0, i0 + i = 0

Rule argument

Additional information

Used simplifier rules: i2b-noteq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j;% ⊦ i = i + j ↔ j = 0;% ⊦ i - j = i0 ↔ i = i0 + j;%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j = j + i;