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Proof step 20 from proof sum-countfun-not-ls (previous step)

i < 0, i0 < j0, 0 < j0, Σ countFun(i0, j0 - 1, intfun) < intfun(j1), i0 ≤ j0, i0 ≤ j1, j1 < j0, 0 ≤ i0, 0 ≤ j0, 0 ≤ j1, i ∈ countFun(i0, j0 - 1, intfun), ∀ 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 ⊦ 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

Proof step 21:
⊦ i0 ∈ countFun(i, j, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = i0)
Proof step 22:
i < 0, i0 < j0, 0 < j0, Σ countFun(i0, j0 - 1, intfun) < intfun(j1), i0 ≤ j0, i0 ≤ j1, j1 < j0, 0 ≤ i0, 0 ≤ j0, 0 ≤ j1, ∃ j1. i0 ≤ j1 ∧ j1 ≤ j0 - 1 ∧ intfun(j1) = i, ∀ 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 ⊦ j0 = j1, ∃ i, j. i0 ≤ i ∧ i < j0 ∧ intfun(i) = intfun(j0) ∧ i0 ≤ j ∧ j < j0 ∧ intfun(j) = intfun(j1) ∧ i ≠ j

Rule argument

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

Additional information

Used terms: intfun, j0 - 1, i0, i