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Proof step 4 from proof int2bits-one-bit (previous step)

i < 0, int2bits(i) = boolvar ' ⊦ i = ~1 ∧ boolvar, i = 0 ∧ ¬ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bits(i) = boolvar ', i < 0 ⊦ i = 0, i = ~1 ∧ boolvar

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;