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Proof step 1 from proof bits2bits-in (conclusion)

⊦ bits ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits)) ↔ boolvar ∈ bits)

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ boolvar, bits0. # bits0 < # bits → true → bits0 ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits0)) ↔ boolvar ∈ bits0) ⊦ bits ≠ [] → (boolvar ∈ bytes2bits(bits2bytes(bits)) ↔ boolvar ∈ bits)

Rule argument

indvar: # bits wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # bits