Is "duplicate" in RPN enough for replacing variable binding in term expressions?

I try to work out some consequences of storing (or "communicating"/"transmitting") a rational number by a term expression using the following operators: $0$, $\mathsf{inc}$, $\mathsf{add}$, $\mathsf{mul}$, $\mathsf{neg}$, and $\mathsf{inv}$. Here $\mathsf{add}$ and $\mathsf{mul}$ are binary operators, $\mathsf{inc}$, $\mathsf{neg}$, and $\mathsf{inv}$ are unary operators, and $0$ is a $0$-ary operator (i.e. a constant). Because I want to be able to also store numbers like $(3^3+3)^3$ efficiently, I need some form of variable binding. I will use the notation $(y:=t(x).f(y))$ to be interpreted as $f(t(x))$ in this question. Now I can store $(3^3+3)^3$ as
$$(c3:=(1+1+1).(x:=((c3c3c3)+c3).(xxx))).$$
If I stick to the operators $0$, $\mathsf{inc}$, $\mathsf{add}$, and $\mathsf{mul}$, this becomes
$$(c3:=\mathsf{inc}(\mathsf{inc}(\mathsf{inc}(0))).(x:=\mathsf{add}(\mathsf{mul}(\mathsf{mul}(c3,c3),c3),c3).\mathsf{mul}(\mathsf{mul}(x,x),x))).$$
Using RPN with a "duplicate" operation written $\mathsf{dup}$ instead of variable binding, this becomes
$$0\ \mathsf{inc}\ \mathsf{inc}\ \mathsf{inc}\ \mathsf{dup}\ \mathsf{dup}\ \mathsf{dup}\ \mathsf{mul}\ \mathsf{mul}\ \mathsf{add}\ \mathsf{dup}\ \mathsf{dup}\ \mathsf{mul}\ \mathsf{mul}.$$

My question is whether it is always possible to replace variable binding by the "duplicate" operation. The binary operations ($\mathsf{add}$ and $\mathsf{mul}$) are associative and commutative, but it seems to me that even this is not enough for ensuring that variable binding can be completely eliminated. Take for example
$$(c2:=(1+1).(x:=(((c2+1)c2)+1).(y:=(xx).((y+c2)*y)))).$$
If I stick to the operators $0$, $\mathsf{inc}$, $\mathsf{add}$, and $\mathsf{mul}$, this becomes
$$(c2:=\mathsf{inc}(\mathsf{inc}(0)).(x:=\mathsf{inc}(\mathsf{mul}(\mathsf{inc}(c2),c2)).(y:=\mathsf{mul}(x,x).\mathsf{mul}(\mathsf{add}(y,c2),y)))).$$
Using RPN with a "store" operation written $\mathsf{sto}(x)$ instead of variable binding, this becomes
$$0\ \mathsf{i

Is “duplicate” in RPN enough for replacing variable binding in term expressions?

No, it is not strong enough. One can eliminate $\mathsf{dup}$ by replacing each sequence of $\mathsf{dup}$ by a sequence of $x$ preceded by one $\mathsf{sto}(x)$ operation. Since the single variable "$x$" is enough to eliminate all $\mathsf{dup}$ operations, any term expression which requires more than one variable to be stored efficiently can't be expressed (efficiently) using only $\mathsf{dup}$.

If the $\mathsf{inv}$ operation is omitted, then this whole intended representation by a term expression can be seen as a special case of the directed acyclic graph representation of a polynomial used in arithmetic circuit complexity. So the RPN stack from the question turns out to be just a representation of a tree. Now a directed acyclic graph can't be "emulated easily" by a tree, and especially a non-planar DAG will be "too difficult" for the $\mathsf{dup}$ operation.

In order to add the $\mathsf{inv}$ operation again, note that the ring of polynomials in $x_1,\dots x_n$ over $\mathbb Z$ is just the free commutative ring with generators $x_1,\dots x_n$. The corresponding free algebra with an $\mathsf{inv}$ operation would be the free commutative regular ring with generators $x_1,\dots x_n$. To be faithful to the canonical representation of the free algebra of a variety (equationally defined collection of algebras), the $\mathsf{inc}$ operation should be replaced by $1$.

StackExchange Computer Science Q#32151, answer score: 2