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Proof step 3 from proof count-i2j-in (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j = 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, j) ↔ i ≤ i0 ∧ i0 ≤ j

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, i, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, i, i2