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Proof step 15 from proof sum-countfun-not-ls (previous step)

¬(intfun(j1) ∈ countFun(i0, j0 - 1, intfun) ∧ (∀ i. i ∈ countFun(i0, j0 - 1, intfun) → 0 ≤ i)), 0 ≤ j1, 0 ≤ j0, 0 ≤ i0, j1 < j0, i0 ≤ j1, i0 ≤ j0, Σ countFun(i0, j0 - 1, intfun) < intfun(j1), 0 < j0, i0 < j0, ∀ i1. i0 ≤ i1 ∧ i1 ≤ j0 → 0 ≤ intfun(i1), ∀ j1, i0, intfun, i, j. i→n(j1) < i→n(j0) → (∃ i1, j2. i0 ≤ i1 ∧ i1 ≤ j1 ∧ intfun(i1) = i ∧ i0 ≤ j2 ∧ j2 ≤ j1 ∧ intfun(j2) = j ∧ i1 ≠ j2) ∧ (∀ i1. i0 ≤ i1 ∧ i1 ≤ j1 → 0 ≤ intfun(i1)) ∧ 0 ≤ i0 → ¬ Σ countFun(i0, j1, intfun) < i + j ⊦ j0 = j1, ∃ i, j. i0 ≤ i ∧ i < j0 ∧ intfun(i) = intfun(j0) ∧ i0 ≤ j ∧ j < j0 ∧ intfun(j) = intfun(j1) ∧ i ≠ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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