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Proof step 1 from proof i2np (conclusion)

0 ≤ i, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ j. 0 ≤ i0 ∧ 0 ≤ j → i→n(i0) * i→n(j) = i→n(i0 * j), 0 ≤ i0 +1, 0 ≤ j ⊦ i→n(i0 +1) * i→n(j) = i→n(i0 +1 * j)
Proof step 12:
∀ j. 0 ≤ i0 ∧ 0 ≤ j → i→n(i0) * i→n(j) = i→n(i0 * j), 0 ≤ i0 -1, 0 ≤ j ⊦ i→n(i0 -1) * i→n(j) = i→n(i0 -1 * j)

Rule argument

indvar: i

Additional information

Used simplifier rules: int generated by 0, +1, -1;;%zero: ⊦ 0 * n = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult