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Proof step 2 from proof bits2bits-in (previous step)

true, ∀ 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 3:
∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits = [], boolvar ∈ bytes2bits(bits2bytes(bits)) ↔ boolvar ∈ bits

Proof step 2 from proof bits2bits-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information