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Proof step 1 from proof byte-length (conclusion)

b2i(by) ≠ ~128 ⊦ # int2bits(b2i(by)) ≤ 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
8 < # int2bits(b2i(by)) ⊦ b2i(by) = ~128

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;