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Proof step 7 from proof int2bits-truein-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j < 0 ⊦ j = 0, true ∈ int2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;