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Proof step 2 from proof byte-succ (previous step)

i < 127 ⊦ 0 ≤ b2i(by) ∧ b2i(by) ≤ i → b2i(i2b(b2i(by) + 1)) = b2i(by) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
b2i(by) < 127, 0 ≤ i, 0 ≤ b2i(by), i < 127, b2i(by) ≤ i ⊦ b2i(i2b(b2i(by) + 1)) = b2i(by) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)