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Proof step 3 from proof bitsxor-bit (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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