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Proof step 14 from proof bits2bytes-app (previous step)

8 ≠ 0, # bits0 ≤ 8, ∃ n0. # bits0 = n0 * 8, # bits + # bits0 ≤ 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ i2b(bits2int(bits + bits0)) ' = bits2bytes(bits) + i2b(bits2int(bits0)) ', bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# bits0 = n * 8, n * 8 ≤ 8, # bits + n * 8 ≤ 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + n * 8 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ i2b(bits2int(bits + bits0)) ' = bits2bytes(bits) + i2b(bits2int(bits0)) ', bits0 = [], bits = []

Proof step 14 from proof bits2bytes-app (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information