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Proof step 22 from proof byte-mult-256 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
~256 ≤ j0, 0 < i, ~256 < j0, j0 < 0 ⊦ i = 0, i2b(256 + j0) = i2b(j0)

Rule argument

Additional information

Used simplifier rules: i2b-noteq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j;% ⊦ i = i + j ↔ j = 0;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;