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Proof step 15 from proof bits2bits-inj (previous step)

# lastn(8, bits) = 8, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + lastn(8, bits) = bits

User applied rule apply induction

And obtained the following new goals

Proof step 16:
# lastn(8, bits) = 8, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ # firstn(# bits - 8, bits) % 8 = 0, bits = [], bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + lastn(8, bits) = bits
Proof step 22:
firstn(# bits - 8, bits) = [] ∨ bytes2bits(bits2bytes(firstn(# bits - 8, bits))) = firstn(# bits - 8, bits), # lastn(8, bits) = 8, # bits % 8 = 0, 8 < # bits, ∀ bits0. # bits0 < # bits → # bits0 % 8 = 0 → bits0 = [] ∨ bytes2bits(bits2bytes(bits0)) = bits0 ⊦ bits = [], bytes2bits(bits2bytes(firstn(# bits - 8, bits))) + lastn(8, bits) = bits

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)