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Proof step 13 from proof count-i2j-in (previous step)

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 = i3 + 1 ↔ i ≤ i0 ∧ i0 ≤ i3 + 1

User applied rule apply induction

And obtained the following new goals

Proof step 14:
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) ⊦ i→n(i3 - i) < i→n(i2), i0 ∈ count-i2j(i, i3) ∨ i0 = i3 + 1 ↔ i ≤ i0 ∧ i0 ≤ i3 + 1
Proof step 16:
i0 ∈ count-i2j(i, i3) ↔ i ≤ i0 ∧ i0 ≤ i3, 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 = i3 + 1 ↔ i ≤ i0 ∧ i0 ≤ i3 + 1

Rule argument

subst: [j0, i1, i0] → [i3, i, i0]

Additional information

Used terms: i3, i, i0