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Proof step 23 from proof byte-law-add-one (previous step)

127 < j0 ∧ j0 < 256, 0 ≤ i0, 0 ≤ j0, j < 0, j0 < 256 ⊦ i2b(j0 + j) = i2b((j0 - 256) + j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
0 < j0, j0 < 256, j < 0, 0 ≤ j0, 0 ≤ i0, 127 < j0 ⊦ i2b(j0 + j) = i2b(j + j0 - 256)

Rule argument

Additional information

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