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Proof step 1 from proof byte-succ (conclusion)
i < 127 ⊦ 0 ≤ b2i(by) ∧ b2i(by) ≤ (* i) → b2i(i2b(b2i(by) + 1)) = b2i(by) + 1
Special step normalize
And obtained the following new goals
Proof step 2
:
i < 127 ⊦ 0 ≤ b2i(by) ∧ b2i(by) ≤ i → b2i(i2b(b2i(by) + 1)) = b2i(by) + 1