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Proof step 1 from proof sum-countfun-not-ls (conclusion)

∃ 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 ⊦ ¬ Σ countFun(i0, j0, intfun) < i + j

User applied rule induction

And obtained the following new goals

Proof step 2:
(∃ 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

Rule argument

indvar: i→n(j0) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j0)