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Proof step 10 from proof countFun-in (previous step)

i ≤ i2 + 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)) ⊦ i0 ∈ intfun(i) ' + countFun(i + 1, i2 + i, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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)

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%in-17: ⊦ i ∈ i0 ' ↔ i = i0; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]