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Proof step 17 from proof sum-countfun-not-ls (previous step)

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 ⊦ intfun(j1) ∈ countFun(i0, j0 - 1, intfun), j0 = j1, ∃ i, j. i0 ≤ i ∧ i < j0 ∧ intfun(i) = intfun(j0) ∧ i0 ≤ j ∧ j < j0 ∧ intfun(j) = intfun(j1) ∧ i ≠ j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: countFun-in-01 seq: ⊦ i ≤ i0 ∧ i0 ≤ j → intfun(i0) ∈ countFun(i, j, intfun) subst: [j, i, i0, intfun] → [j0 - 1, i0, j1, intfun] flags: rewrite, current

Additional information

Used simplifier rules: sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j0 - 1, i0, j1, intfun