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Proof step 2 from proof bitsand-idempotent (previous step)

⊦ bits = [], bitsand(bits, bits) = bits

User applied rule induction

And obtained the following new goals

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

Rule argument

indvar: # bits wfpred: <

Additional information

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