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Proof step 3 from proof and-one (previous step)

¬(boolvar ' ≠ [] ∧ boolvar0 ' = true ') ⊦ bitsand(boolvar ', boolvar0 ') = (boolvar ∧ boolvar0)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bitsand(boolvar ', false ') = false '

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];