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Proof step 4 from proof byte-law-add-pos (previous step)

¬(0 ≤ i ∧ 127 < i % 256), 0 ≤ i, 0 ≤ j ⊦ i2b(i + j) = i2b(b2i(i2b(i)) + j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 ≤ j, i % 256 ≤ 127 ⊦ i2b(i + j) = i2b(i % 256 + j)

Rule argument

Additional information

Used simplifier rules: bib-pos-low: ⊦ 0 ≤ i ∧ i % 256 ≤ 127 → b2i(i2b(i)) = i % 256;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;