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Proof step 10 from proof bits2bytes-len (previous step)

8 < # bits, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ bits = [], # bits2bytes(firstn(# bits - 8, bits)) = (# bits - 1) / 8

User applied rule apply induction

And obtained the following new goals

Proof step 11:
firstn(# bits - 8, bits) = [] ∨ # bits2bytes(firstn(# bits - 8, bits)) = (# firstn(# bits - 8, bits) - 1) / 8 + 1, 8 < # bits, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ # bits2bytes(bits0) = (# bits0 - 1) / 8 + 1 ⊦ bits = [], # bits2bytes(firstn(# bits - 8, bits)) = (# bits - 1) / 8

Rule argument

subst: [bits0] → [firstn(# bits - 8, bits)]

Additional information

Used simplifier rules: ⊦ # firstn(n, bits) = n;% ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m + n ≤ m ↔ n = 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: firstn(# bits - 8, bits)