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Proof step 16 from proof countFun-in (previous step)

intfun(i) ∈ countFun(i + 1, i2 + i, intfun) ↔ (∃ j. i < j ∧ j ≤ i2 + i ∧ intfun(j) = intfun(i)), 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)) ⊦ ∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = intfun(i)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i

Additional information

Used simplifier rules: ler: ⊦ i ≤ i; from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ 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: i