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Proof step 4 from proof int2bits-truein-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ j = 0, true ∈ int2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;% ⊦ ¬ i ≤ j ↔ j < i;