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Proof step 6 from proof bits2bytes-len (previous step)

¬ # bits ≤ 8, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ # bits2bytes(bits) = (# bits - 1) / 8 + 1, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
8 < # bits, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ bits = [], # bits2bytes(bits) = (# bits - 1) / 8 + 1

Rule argument

Additional information

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