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Proof step 19 from proof bits2bytes-fill (previous step)

8 ≤ # mklist(bits.first, 8 - # bits % 8) + # bits ∧ # mklist(bits.first, 8 - # bits % 8) ≤ 8, ((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits ⊦ bits = [], # bits % 8 = 0, bits2bytes(bits) = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)) + bits2bytes(restn(8 - # mklist(bits.first, 8 - # bits % 8), bits)), 8 + # bits = # bits % 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
((8 - # bits % 8) + # bits) % 8 = 0, 8 < (8 - # bits % 8) + # bits ⊦ 8 + # bits = # bits % 8, bits = [], # bits % 8 = 0, bits2bytes(bits) = bits2bytes(firstn(8, mklist(bits.first, 8 - # bits % 8) + bits)) + bits2bytes(restn(8 - (8 - # bits % 8), bits))

Rule argument

Additional information

Used simplifier rules: ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m + n;%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;%n ≠ 0 ⊦ m % n ≤ n;% ⊦ m < n → m ≤ n;% ⊦ m + n = n + m;% ⊦ # mklist(boolvar, n) = n;