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Proof step 31 from proof bits2bytes-app (previous step)

8 < # bits0, # bits0 % 8 = 0, 8 < # bits + # bits0, 8 ≤ # bits0, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ bits2bytes(firstn(# bits + (# bits0 - 8), bits + bits0)) + i2b(bits2int(lastn(8, bits0))) ' = bits2bytes(bits) + (bits2bytes(firstn(# bits0 - 8, bits0)) + i2b(bits2int(lastn(8, bits0)))), bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
# bits0 % 8 = 0, 8 < # bits0, 8 < # bits + # bits0, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ bits0 = [], bits = [], bits2bytes(firstn(# bits + (# bits0 - 8), bits + bits0)) = bits2bytes(bits) + bits2bytes(firstn(# bits0 - 8, bits0))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ bytes + by = bytes + by ';