Let A = {-2, -1, 0, 1, 2, 3, 4}. Let R be a r | vedclass

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(C) Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$ and $xRy \iff 2x + y \le 2$.For each $x \in A$, we find $y \in A$ such that $y \le 2 - 2x$:- If $x = -2$, $y

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$33$

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(C) Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$ and $xRy \iff 2x + y \le 2$.For each $x \in A$, we find $y \in A$ such that $y \le 2 - 2x$:- If $x = -2$, $y \le 6 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).- If $x = -1$, $y \le 4 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).- If $x = 0$, $y \le 2 \implies y \in \{-2, -1, 0, 1, 2\}$ ($5$ elements).- If $x = 1$, $y \le 0 \implies y \in \{-2, -1, 0\}$ ($3$ elements).- If $x = 2$, $y \le -2 \implies y \in \{-2\}$ ($1$ element).- If $x = 3$, $y \le -4 \implies$ no $y \in A$.- If $x = 4$, $y \le -6 \implies$ no $y \in A$.Total elements $l = 7 + 7 + 5 + 3 + 1 = 23$.For reflexivity, we need $(x, x) \in R$ for all $x \in A$. Check: $(-2, -2), (-1, -1), (0, 0)$ are in $R$. $(1, 1), (2, 2), (3, 3), (4, 4)$ are missing. So $m = 4$.For symmetry, if $(x, y) \in R$, then $(y, x)$ must be in $R$. The elements $(x, y)$ in $R$ where $(y, x) 
otin R$ are: $(3, -2), (4, -2), (2, -1), (2, 0), (3, -1), (4, -1)$. There are $6$ such pairs. So $n = 6$.Thus, $l + m + n = 23 + 4 + 6 = 33$.

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