$\color{red}{\text{Detailed Video Solution:}}$ Detailed Video Solution with Complete Analysis
Option A Explanation:
Option A says "For every element $x$, there is an element $y$ such that $\mathrm{P}(x, y)$ is true". This is false, and the counterexample is the element $x=$ 4. (Since there is No element $y$ such that $\mathrm{P}(4, y)$ is true.)
Option B Explanation:
Option B says "For every element $y$, there is an element $x$ such that $\mathrm{P}(x, y)$ is true". This is true because:
- For $y=1$, we have $x=1$ or 3 . (Since $P(1,1)=T, P(3,1)=T)$
- For $y=2$, we have $x=2$ or 3. (Since $\mathrm{P}(2,2)=\mathrm{T}, \mathrm{P}(3,2)=\mathrm{T}$ )
- For $y=3$, we have $x=2$ or 3 . (Since $P(2,3)=T, P(3,3)=T$ )
- For $y=4$, we have $x=3$. $($ Since $P(3,4)=T)$
Option C Explanation:
Option C says "There is some element $x$, such that for all elements $y$, $P(x, y)$ is true".. This is true because of $x=3$.
For $x=3$, we have $\mathrm{P}(3,1)=\mathrm{T}, \mathrm{P}(3,2)=\mathrm{T}, \mathrm{P}(3,3)=\mathrm{T}, \mathrm{P}(3,4)=\mathrm{T}$.
Option D Explanation:
Option D says "There is some element $y$, such that for all elements $x$, $\mathrm{P}(x, y)$ is true". This is false there is no such element $y$.
- For $y=1$, we have $P(2,1)=F$.
- For $y=2$, we have $\mathrm{P}(1,2)=\mathrm{F}$.
- For $y=3$, we have $P(1,3)=F$.
- For $y=4$, we have $\mathrm{P}(1,4)=F.$
$\color{red}{\text{Detailed Video Solution:}}$ Detailed Video Solution with Complete Analysis
