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Proof step 1 from proof add-ones-grlen-02 (conclusion)

⊦ ¬ false ∈ bits ∧ # bits0 ≤ # bits ∧ ¬ bits0.first ∧ bits0 ≠ [] → bitsand(bits, bits0) = bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# bits0 ≤ # bits ⊦ bits0 = [], bitsand(bits, bits0) = bits0, false ∈ bits, bits0.first

Rule argument

Additional information