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Proof step 5 from proof natpos (previous step)

i→n(i0 * j + 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 6:
0 ≤ i, 0 ≤ j ⊦ i→n(i) + i→n(j) = i→n(i + j)
Proof step 7:
i→n(i0 * j) + i→n(j0) = m0 * i→n(j) + n0, n0 < i→n(j), j0 < j, 0 < j, 0 ≤ i0, 0 ≤ i0 * j, 0 ≤ j0 ⊦ j = 0, m0 = i→n(i0)

Rule argument

spec: <toplevel> name: add seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) + i→n(j) = i→n(i + j) subst: [j, i] → [j0, i0 * j] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0, i0 * j