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Proof step 12 from proof natpos (previous step)

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)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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