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Proof step 2 from proof bits2bytes-rec-02 (previous step)

true, ∀ bits0. # bits0 < # bits → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ # bits % 8 = 0 ∧ 8 < # bits → bits2bytes(bits) = bits2bytes(firstn(8, bits)) + bits2bytes(restn(8, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits2bytes(bits) = bits2bytes(firstn(8, bits)) + bits2bytes(restn(8, bits)), bits = []

Rule argument

Additional information

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