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Proof step 4 from proof count-i2j-in (previous step)

j = i2 + i, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, j) ↔ i ≤ i0 ∧ i0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information