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Proof step 38 from proof bits2bytes-app (previous step)

8 ≠ 0, ∃ n0. # bits0 = n0 * 8, 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(bits + firstn(# bits0 - 8, bits0)) = bits2bytes(bits) + bits2bytes(firstn(# bits0 - 8, bits0)), (# bits0 - 8) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
# bits0 = n * 8, 8 < # bits + n * 8, 8 < n * 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + n * 8 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ (n * 8 - 8) % 8 = 0, bits2bytes(bits + firstn(n * 8 - 8, bits0)) = bits2bytes(bits) + bits2bytes(firstn(n * 8 - 8, bits0)), bits = [], bits0 = [], n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ m < n → n ≠ 0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)