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Proof step 6 from proof bits2bytes-app (previous step)

¬ 8 < #(bits + bits0), # bits0 % 8 = 0, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ bits2bytes(bits + bits0) = bits2bytes(bits) + bits2bytes(bits0), bits0 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits0 % 8 = 0, # bits + # bits0 ≤ 8, ∀ bits1, bits2. # bits2 + # bits1 < # bits + # bits0 → # bits1 % 8 = 0 → bits2 = [] ∨ bits1 = [] ∨ bits2bytes(bits2 + bits1) = bits2bytes(bits2) + bits2bytes(bits1) ⊦ bits2bytes(bits + bits0) = bits2bytes(bits) + bits2bytes(bits0), bits0 = [], bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;