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Proof step 3 from proof bitsand-idempotent (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information