Lambda Calculus syntax: multiplication is or isn't implied?

Question

If I have this

$$(\lambda xy.xy)\; 1 \; 2$$

does this beta reduce to ($1 \; 2$) or is multiplication finally implied ($1 \cdot 2$), i.e., $2$ -- the former being just $1$ then $2$ beside it, not interacting?

Answer 1

There is no multiplication here, since $1, 2$ are just letters. So yes it reducts to the word $"12"$. Don't write the braces (they are just here for convenience to help us humans parse what we read but can be omited) , they vanish upon $\lambda$-reduction.

$$\begin{array}{ccccc} (\lambda xy.xy)\; 1 \; 2 & := & \big(\lambda x.(\lambda y.xy) \big)\;1 \; 2 & \longrightarrow & (\lambda y.1y) \; 2 & \longrightarrow & 12 & \end{array}$$