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Proof step 4 from proof bits2bytes-fill (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ bits2bytes(bits) = bits2bytes(mklist(bits.first, 8 - # bits % 8) + bits), # bits % 8 = 0, bits = [], ((8 - # bits % 8) + # bits) % 8 = 0 ∧ ¬ # bits ≤ # bits % 8

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n;%n ≠ 0 ⊦ m % n ≤ n;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;