Question Why are functions of a single variable specified with a tuple argument $(x)$ as in $f(x)$, when this pattern suggests they should be specified without parenthesis as $fx$?
This pattern
(1): $$ X = \{a\} $$ $$X^n = X \times ... \times X$$ $$ X^0 = \{()\}, X_0() = T, () \in X^0 $$ $$ \mathbf {X^1 = \{a\}, X_1(a) = T , a \in X^1} $$ $$ X^2 = \{(a,a)\}, X_2(a,a) = T, (a,a) \in X^2 $$
where $X_n : X^n \rightarrow \{T,F\}$
The pattern continues $X^n, n>2$. If we follow the pattern, $X^1$ would be elements of tuples ($f :\mathbb N \rightarrow X$):
(2): $$ \mathbf{X^1 = \{(a)\}, X_1(a) = T , (a) \in X^1} $$ With this, the predicate is the relation's symbol $X_1$ and the object is the ordered tuple $(x_0, ..., x_n)$. This case is not true, which means the predicate may have either a tuple or individual as it's object.
To be consistent with the pattern, it would seem the relation $X$ does not take an ordered pair $(a)$ as an argument and is without parenthesis: $$ \mathbf {X^1 = \{a\}, X_1 a = T , a \in X^1} $$ but I know an ordered pair $(a)$ is typically given as an argument. Given this, it appears the parenthesis is syntactical and does not form an ordered tuple.
If the pattern (2) is followed the notation for composition becomes more clear, $f : X^1 \rightarrow X^1$, $f = \{((a),(a))\}$ then $f(a) = (a)$ and the composition is just $f...f(a)$. Otherwise with pattern (1) composition requires the syntactical brackets $f(...f(a))$. Again, with a tuple, $g : X^3 \rightarrow X^3$ would be composed $g(g(x))$ with these extra syntactical parenthesis vs $g.g(x)$ which is confusing.