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Proof step 16 from proof count-i2j-in (previous step)

i0 ∈ count-i2j(i, i3) ↔ i ≤ i0 ∧ i0 ≤ i3, i2 + i = i3 + 1, 0 ≤ i2, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i3) ∨ i0 = i3 + 1 ↔ i ≤ i0 ∧ i0 ≤ i3 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
i2 + i = i3 + 1, i0 ∈ count-i2j(i, i3) ↔ i ≤ i0 ∧ i0 ≤ i3, 0 ≤ i2, i ≤ i0 ∧ i0 ≤ i3 ∨ i0 = i3 + 1, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i ≤ i0 ∧ i0 ≤ i3 + 1

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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs