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Proof step 4 from proof ubyte-pos-03 (previous step)

¬ 0 ≤ b2i(i2b(i)), i < 256, 0 ≤ i ⊦ ub2i(i2b(i)) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b2i(i2b(i)) < 0, 0 ≤ i, i < 256 ⊦ 256 + b2i(i2b(i)) = i

Rule argument

Additional information

Used simplifier rules: ubyte-neg: ⊦ b2i(by) < 0 → ub2i(by) = 256 + b2i(by);% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;