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Proof step 8 from proof natpos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i→n(i) * i→n(j) = i→n(i * j)
Proof step 10:
i→n(i0) * i→n(j) + i→n(j0) = m0 * i→n(j) + n0, n0 < i→n(j), j0 < j, 0 < j, 0 ≤ i0, 0 ≤ j0 ⊦ j = 0, m0 = i→n(i0)

Rule argument

spec: <toplevel> name: i2n seq: ⊦ i→n(i) * i→n(j) = i→n(i * j) subst: [j, i] → [j, i0] flags: <rotated> <all paths>

Additional information

Used terms: j, i0