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Proof step 8 from proof count-i2j-in (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ j < i → count-i2j(i, j) = []

Rule argument

spec: inst: name: counti2j-empty seq: ⊦ j < i → count-i2j(i, j) = [] subst: [j, i] → [i2 + i, i] flags: rewrite, current

Additional information

Used simplifier rules: in-empty: ⊦ ¬ i ∈ []; from specification list[intlista]%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs Used terms: i2 + i, i