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Proof step 12 from proof bits2bits-inj (previous step)

# lastn(8, bits) ≠ 8, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + byte2bits(i2b(bits2int(lastn(8, bits)))) = bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ # lastn(8, bits) = 8, bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + byte2bits(i2b(bits2int(lastn(8, bits)))) = bits, bits = []

Proof step 12 from proof bits2bits-inj (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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