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Proof step 2 from proof bits2uint-zeros (previous step)

⊦ bits2uint(bits) = 0, true ∈ bits

User applied rule induction

And obtained the following new goals

Proof step 3:
true, ∀ bits0. # bits0 < # bits → true → bits2uint(bits0) = 0 ∨ true ∈ bits0 ⊦ bits2uint(bits) = 0 ∨ true ∈ bits

Rule argument

indvar: # bits wfpred: <

Additional information

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