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Proof step 2 from proof two-add-square (previous step)

0 ≤ i ⊦ (i * (i - 1)) / 2 + i = (i * (i + 1)) / 2 ↔ true

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → i / j + i0 = (i + i0 * j) / j

Rule argument

spec: <toplevel> name: div-mult-pos-rew seq: ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → i / j + i0 = (i + i0 * j) / j subst: [i0, j, i] → [i, 2, i * (i - 1)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i * j + i * j0 = i * (j + j0); (s)% ⊦ i + (j - i0) = i + j - i0; (s)% ⊦ (* i) + i0 - (* j) = (i - j) + i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 0 ≤ i → (0 ≤ i * j ↔ i = 0 ∨ 0 ≤ j); (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) Used terms: i, 2, i * (i - 1)