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Proof step 5 from proof count-i2j-in (previous step)

∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i2 + i) ↔ i ≤ i0 ∧ i0 ≤ i2 + i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i ≤ j → count-i2j(i, j) = count-i2j(i, j - 1) + j
Proof step 7:
¬ i ≤ i2 + i, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i2 + i) ↔ i ≤ i0 ∧ i0 ≤ i2 + i
Proof step 10:
i ≤ i2 + i, ∀ j0, i1, i0. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ count-i2j(i1, j0) ↔ i1 ≤ i0 ∧ i0 ≤ j0) ⊦ i0 ∈ count-i2j(i, i2 + i - 1) + (i2 + i) ↔ i ≤ i0 ∧ i0 ≤ i2 + i

Rule argument

spec: inst: name: counti2j-rec seq: ⊦ i ≤ j → count-i2j(i, j) = count-i2j(i, j - 1) + j subst: [j, i] → [i2 + i, i] flags: rewrite, current

Additional information

Used terms: i2 + i, i