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Proof step 35 from proof sum-countfun-not-ls (previous step)

Σ countFun(i0, j0 - 1, intfun) < intfun(i1), 0 ≤ j0, 0 ≤ i1, 0 ≤ i0, i0 ≤ j0, i1 < j0, i0 < j0, 0 < j0, i0 ≤ i1, i0 ≤ j, j < j0, intfun(j) < 0, 0 ≤ j, ∀ 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 ⊦ i1 = j0, ∃ i, j. i0 ≤ i ∧ i < j0 ∧ intfun(i) = intfun(i1) ∧ i0 ≤ j ∧ j < j0 ∧ intfun(j) = intfun(j0) ∧ i ≠ j

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 14 terms: j

Additional information

Used simplifier rules: lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j