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Proof step 13 from proof ubyte-inj (previous step)

¬ b2i(by0) < 0, b2i(by) < 0, 256 + b2i(by) = ub2i(by0) ⊦ by = by0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
256 + b2i(by) = b2i(by0), 0 ≤ b2i(by0), b2i(by) < 0 ⊦ by = by0

Rule argument

Additional information

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