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Proof step 20 from proof countFun-in (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: j

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs Used terms: j