Lemma·L14
Associativity of multiplication
Grouping does not matter when multiplying.
for all
, with
from D028.
In words
Multiplying m by n and then by k gives the same result as multiplying m by the product of n and k, for all natural numbers m, n, k.
Never needed: F03 · F04 · F05 · F06 · F08 · F10 · F11 · F12 · F13 · A02 · A03 · A04 · A05 · A07 · A08 (computed from the citation graph, not asserted).
Proof
- 1Fix and ; induct on (T05, Separation).
- 2Base: by D028 applied on each side.
- 3Step: assume . Then by D028, the hypothesis with substitutivity, distributivity read right to left, and D028 once more. T05 closes the induction.
∎
Remarks
The proof rides on distributivity: pulling one factor of the product outward is exactly a distributive regrouping. With commutativity, all bracketings and orderings of a product of naturals agree, which the notation silently assumes from here on, e.g. in transitivity of divisibility.