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Proof step 1 from proof bitsand-zero-in (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → (false ∈ bitsand(bits, bits0) ↔ false ∈ bits ∨ false ∈ bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
false ∈ bits ∨ false ∈ bits0 ⊦ bits0 = [], bits = [], false ∈ bitsand(bits, bits0)

Rule argument

Additional information

Used simplifier rules: and-onesin: ⊦ ¬ false ∈ bits ∧ bits ≠ [] ∧ bits0 ≠ [] → (boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0);