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Proof step 18 from proof count-i2j-no-dups (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 19:
i1 + i = i2 + 1, 0 ≤ i1, dups(count-i2j(i, i2)), ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → dups(count-i2j(i0, j0)) → false ⊦ i→n(i2 - i) < i→n(i1), i2 + 1 ∈ count-i2j(i, i2)

Rule argument

subst: [j0, i0] → [i2, i]

Additional information

Used terms: i2, i