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Proof step 10 from proof natpos (previous step)

i→n(i0) * i→n(j) + i→n(j0) = m0 * i→n(j) + n0, 0 ≤ i0, 0 < j, 0 ≤ j0, j0 < j, n0 < i→n(j) ⊦ j = 0, m0 = i→n(i0)

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 11:
⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1)
Proof step 12:
n0 < i→n(j) ∧ i→n(j0) < i→n(j) → (m0 * i→n(j) + n0 = i→n(i0) * i→n(j) + i→n(j0) ↔ m0 = i→n(i0) ∧ n0 = i→n(j0)), i→n(i0) * i→n(j) + i→n(j0) = m0 * i→n(j) + n0, 0 ≤ i0, 0 < j, 0 ≤ j0, j0 < j, n0 < i→n(j) ⊦ j = 0, m0 = i→n(i0)

Rule argument

spec: nat-mult inst: name: modeq seq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1) subst: [n0, n, n1, m, m0] → [n0, i→n(j), i→n(j0), m0, i→n(i0)] flags: []

Additional information

Used terms: n0, i→n(j), i→n(j0), m0, i→n(i0)