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Proof step 2 from proof int2bits-truein-01 (previous step)

true ∈ int2bits(j).rest ⊦ j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 ^ i + j = 0, true ∈ int2bits(j).rest, 0 ≤ i ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ ¬ boolvar ∈ [];% ⊦ boolvar '.rest = [];% ⊦ ~ i = j ↔ i + j = 0;