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Proof step 17 from proof countFun-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
j ≤ i2 + i, i < j, intfun(j) ∈ countFun(i + 1, i2 + i, intfun), 0 ≤ i2, 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)) ⊦ intfun(j) = intfun(i), ∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = intfun(j)

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs