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Proof step 2 from proof sum-countfun-not-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ j1, 0 ≤ i1, 0 ≤ i0, i0 ≤ j0, j1 ≤ j0, i0 ≤ j1, i1 ≤ j0, i0 ≤ i1, Σ countFun(i0, j0, intfun) < intfun(i1) + intfun(j1), 0 ≤ 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 ⊦ i1 = j1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs