Given the set S = 2, 3, 4, 5 and the operation defined as x y = (x x y) 6, find the truth set of a a = a and evaluate P = (4 5) 3.

We need to calculate $x \otimes y = (x \times y) \pmod 6$ for the given values.

• $2 \otimes 3 = (2 \times 3) \pmod 6 = 6 \pmod 6 = 0$
• $2 \otimes 5 = (2 \times 5) \pmod 6 = 10 \pmod 6 = 4$
• $3 \otimes 4 = (3 \times 4) \pmod 6 = 12 \pmod 6 = 0$
• $4 \otimes 2 = (4 \times 2) \pmod 6 = 8 \pmod 6 = 2$
• $4 \otimes 4 = (4 \times 4) \pmod 6 = 16 \pmod 6 = 4$
• $5 \otimes 3 = (5 \times 3) \pmod 6 = 15 \pmod 6 = 3$

The completed table is:

$$\begin{array}{|c|c|c|c|c|} \hline \otimes & 2 & 3 & 4 & 5 \\ \hline 2 & 4 & 0 & 2 & 4 \\ \hline 3 & 0 & 0 & 0 & 3 \\ \hline 4 & 2 & 0 & 4 & 2 \\ \hline 5 & 4 & 3 & 2 & 1 \\ \hline \end{array}$$

We check each value of $a$ from the set $\{2, 3, 4, 5\}$ using the completed table:

• For $a=2$: $2 \otimes 2 = 4$. Since $4 \neq 2$, $a=2$ is not a solution.
• For $a=3$: $3 \otimes 3 = 0$. Since $0 \neq 3$, $a=3$ is not a solution.
• For $a=4$: $4 \otimes 4 = 4$. Since $4 = 4$, $a=4$ is a solution.
• For $a=5$: $5 \otimes 5 = 1$. Since $1 \neq 5$, $a=5$ is not a solution.

The truth set is $\boxed{\{4\}}$.

First, evaluate the expression inside the parentheses: $4 \otimes 5$.
From the table, $4 \otimes 5 = 2$.
Now substitute this value back into the expression for $P$:
$P = 2 \otimes 3$.
From the table, $2 \otimes 3 = 0$.

Therefore, $P = \boxed{0}$.