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

⊦ bits = [], sgn(bytes2bits(bits2bytes(bits))) = sgn(bits)

User applied rule induction

And obtained the following new goals

Proof step 3:
true, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ sgn(bytes2bits(bits2bytes(bits0))) = sgn(bits0) ⊦ bits = [] ∨ sgn(bytes2bits(bits2bytes(bits))) = sgn(bits)

Rule argument

indvar: # bits wfpred: <

Additional information

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