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Proof step 5 from proof bits2bits-in (previous step)

¬ 8 < # bits, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ bytes2bits(bits2bytes(bits)) ↔ boolvar ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# bits ≤ 8, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ byte2bits(i2b(bits2int(bits))) ↔ boolvar ∈ bits

Rule argument

Additional information

Used simplifier rules: bits2bytes-base: ⊦ # bits ≤ 8 → bits2bytes(bits) = i2b(bits2int(bits)) ';%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ ¬ m < n ↔ n ≤ m;