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Proof step 2 from proof three (previous step)

i * j = 3, 0 < i, 1 < i ⊦ i = 3 ∧ j = 1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
i * i0 + j0 = i1 * j + j1, 0 ≤ j1, j1 < j, 0 ≤ j0, j0 < j, 0 ≤ i0, i0 ≤ j, 0 ≤ i, i < i1 ⊦
Proof step 4:
i * j = 3, 0 < i, 1 < i ⊦ 1 * 3 + 0 = j * i + 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 3 ∧ 3 ≤ i ∧ 0 ≤ 1 ∧ 1 < j, i = 3 ∧ j = 1

Rule argument

spec: int-mult inst: name: modneqp seq: i * i0 + j0 = i1 * j + j1, 0 ≤ j1, j1 < j, 0 ≤ j0, j0 < j, 0 ≤ i0, i0 ≤ j, 0 ≤ i, i < i1 ⊦ subst: [i, i0, j0, i1, j, j1] → [1, 3, 0, j, i, 0] flags: current, []

Additional information

Used terms: 1, 3, 0, j, i, 0