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Proof step 18 from proof int2bits-truein-01 (previous step)

0 ≤ j ⊦ j = 0, true ∈ uint2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 < j ⊦ j = 0, true ∈ uint2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);