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Proof step 14 from proof bitsor-sgn-in (previous step)

bits0 ≠ [] ∧ true ' = true ', true ∈ bits ⊦ bits = [], bits0 = [], boolvar ∈ true ' ↔ boolvar ∈ bitsor(bits, bits0), false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
true ∈ bits ⊦ bits0 = [], bits = [], boolvar ↔ boolvar ∈ bitsor(bits, bits0), false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;