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Proof step 7 from proof count-i2j-in (previous step)

¬ i ≤ i2 + i, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i2 + i) ↔ i ≤ i0 ∧ i0 ≤ i2 + i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i2 < 0, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i2 + i) ↔ i ≤ i0 ∧ i0 ≤ i2 + i

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (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: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs