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Proof step 6 from proof count-i2j-no-dups (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i ≤ j → count-i2j(i, j) = count-i2j(i, j - 1) + j
Proof step 8:
¬ i ≤ i1 + i, dups(count-i2j(i, i1 + i)), ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦
Proof step 11:
i ≤ i1 + i, dups(count-i2j(i, i1 + i - 1) + (i1 + i)), ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦

Rule argument

spec: inst: name: counti2j-rec seq: ⊦ i ≤ j → count-i2j(i, j) = count-i2j(i, j - 1) + j subst: [j, i] → [i1 + i, i] flags: rewrite, current

Additional information

Used terms: i1 + i, i