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Proof step 11 from proof count-i2j-in (previous step)

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, i2 + i - 1) ∨ i0 = i2 + i ↔ i ≤ i0 ∧ i0 ≤ i2 + i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
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 = i2 + i ↔ i ≤ i0 ∧ i0 ≤ i2 + i

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i2 + i, 1, i3] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i2 + i, 1, i3