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Proof step 1 from proof byte-127 (conclusion)

⊦ 126 < b2i(by) ↔ b2i(by) = 127

User applied rule case distinction

And obtained the following new goals

Proof step 2:
126 < b2i(by) ⊦ b2i(by) = 127

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)