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Proof step 1 from proof int2bits-truein (conclusion)

⊦ true ∈ int2bits(j) ↔ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ j = 0, true ∈ int2bits(j)

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;