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Proof step 11 from proof bits2bits-in (previous step)

8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ boolvar ∈ bytes2bits(bits2bytes(firstn(# bits - 8, bits))) ∨ boolvar ∈ byte2bits(i2b(bits2int(lastn(8, bits)))) ↔ boolvar ∈ bits, bits = []

User applied rule apply induction

And obtained the following new goals

Rule argument

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

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), boolvar