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Proof step 12 from proof count-i2j-no-dups (previous step)

dups(count-i2j(i, i1 + i - 1) + (i1 + i)'), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
i1 + i = i2 + 1, dups(count-i2j(i, i2) + (i1 + i)'), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i1 + i, 1, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1 + i, 1, i2