Check whether the following statement is true or not.
If $x, y \in \mathbb{Z}$ are such that $x$ and $y$ are odd,then $xy$ is odd.
If $x, y \in \mathbb{Z}$ are such that $x$ and $y$ are odd,then $xy$ is odd.
(A) Let $p: x, y \in \mathbb{Z}$ such that $x$ and $y$ are odd.
Let $q: xy$ is odd.
To check the validity of the given statement,we assume that if $p$ is true,then $q$ is true.
If $p$ is true,then $x$ and $y$ are odd integers.
We can write $x = 2m + 1$ and $y = 2n + 1$ for some integers $m, n \in \mathbb{Z}$.
Then,$xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$.
Since $2mn + m + n$ is an integer,$xy$ is of the form $2k + 1$,which means $xy$ is odd.
Therefore,the statement is true.
Let $q: xy$ is odd.
To check the validity of the given statement,we assume that if $p$ is true,then $q$ is true.
If $p$ is true,then $x$ and $y$ are odd integers.
We can write $x = 2m + 1$ and $y = 2n + 1$ for some integers $m, n \in \mathbb{Z}$.
Then,$xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$.
Since $2mn + m + n$ is an integer,$xy$ is of the form $2k + 1$,which means $xy$ is odd.
Therefore,the statement is true.
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$(III)$ $(p \to q) \to r \equiv (p \to r) \land (\sim q \to r)$
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