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Proof step 13 from proof countFun-in (previous step)

i0 ∈ countFun(i + 1, i2 + i, intfun) ↔ (∃ j. i < j ∧ j ≤ i2 + i ∧ intfun(j) = i0), 0 ≤ i2, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ countFun(i1, j0, intfun) ↔ (∃ j1. i1 ≤ j1 ∧ j1 ≤ j0 ∧ intfun(j1) = i0)) ⊦ i0 = intfun(i) ∨ i0 ∈ countFun(i + 1, i2 + i, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = i0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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