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Proof step 4 from proof square-lez-01 (previous step)

i = 0 ∨ i = 0 ∨ 0 ≤ i ∧ i ≤ 0 ∨ i ≤ 0 ∧ 0 ≤ i ⊦ i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i ∧ i < 0 ∨ i < 0 ∧ 0 < i ⊦ i = 0

Rule argument

Additional information

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