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Proof step 1 from proof int2bits-eq (conclusion)

⊦ int2bits(j) = false ' ↔ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
int2bits(j) = false ' ⊦ j = 0

Rule argument

Additional information

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