Recall that an algebraic theory (in the sense of Lawvere) is a category $\mathcal{C}$ which is closed under taking finite products, and whose set of objects can be identified with the set $\mathrm{Ob}(\mathcal{C})\cong\{T^0,T^1,\ldots\}$, where $T^i=T^1\times T^1\times\cdots\times T^1$ is the $i$-fold product of $T^1$. We denote $T^1$ by $T$.

For any small category $\mathcal{C}$, say that a $\mathcal{C}$-model is a functor $\mathcal{C}\to\mathbf {Set}$ that preserves all finite products. Denote the category of $\mathcal{C}$-models (and natural transformations between them) by $\mathbf{Mod}(\mathcal{C})$.

Let $n\in\mathbb{N}$ be a natural number, let $\mathcal{C}$ be an algebraic theory, and let $\mathcal{C}_{\leq n}\subseteq\mathcal{C}$ denote the full subcategory spanned by the objects $\{T^0,\ldots,T^n\}$. The following is surely not the best definition, but it's expedient, and I hope it's equivalent to the usual one. I'll say a Lawvere theory $\mathcal{C}$ is $n$-truncated if the induced functor $$\mathbf{Mod}(\mathcal{C})\to\mathbf{Mod}(\mathcal{C}_{\leq n})$$ is an equivalence of categories.

Let $R$ denote the algebraic theory whose morphisms $T^n\to T$ are the smooth functions ${\mathbb R}^n\to{\mathbb R}$. This of course defines the set of morphisms $T^n\to T^m$ for any $m$. Endow $R$ with the usual formula for composing smooth functions.

To me it would be quite surprising if $R$ were 2-truncated. But I've never heard of a 3-ary function $\mathbb{R}^3\to\mathbb{R}$ that wasn't constructed from a combination of 0-, 1-, and 2-ary functions.

Question: Can you prove that $R$ is not 2-truncated? More interestingly, can you name a 3-ary function that isn't constructible by a combination of 0-, 1-, and 2-ary functions?

(Secondary "question": A nicer definition of $n$-truncation would be appreciated.)