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Proof step 4 from proof count-i2j-no-dups (previous step)

dups(count-i2j(i, j)), ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → dups(count-i2j(i0, j0)) → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
j = i1 + i, dups(count-i2j(i, j)), ∀ 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] → [j, i, i1] flags:

Additional information

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