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

⊦ bits = [], 2 ^ n→i(# bits) = bits2uint(bits) + 1, false ∈ bits

User applied rule induction

And obtained the following new goals

Proof step 3:
true, ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + 1 ∨ false ∈ bits0 ⊦ bits = [] ∨ 2 ^ n→i(# bits) = bits2uint(bits) + 1 ∨ false ∈ bits

Rule argument

indvar: # bits wfpred: <

Additional information

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