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Proof step 7 from proof sum-countfun-not-ls (previous step)

0 ≤ j0, 0 ≤ j1, 0 ≤ i1, 0 ≤ i0, i0 ≤ j0, j1 ≤ j0, i0 ≤ j1, i1 ≤ j0, i0 ≤ i1, Σ countFun(i0, j0 - 1, intfun) + intfun(j0) < intfun(i1) + intfun(j1), ∀ 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 ⊦ (∃ i2, j2. i0 ≤ i2 ∧ i2 ≤ j0 - 1 ∧ intfun(i2) = intfun(i1) ∧ i0 ≤ j2 ∧ j2 ≤ j0 - 1 ∧ intfun(j2) = intfun(j1) ∧ i2 ≠ j2) ∧ (∀ i1. i0 ≤ i1 ∧ i1 ≤ j0 - 1 → 0 ≤ intfun(i1)) ∧ 0 ≤ i0, i1 = j1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs