We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
We learned expression of deduce, i.e. =>.
But now I dont have capable reason for I agree I represent True if assumption is False.
Anybady there having to explain reason for its deduce?
Best regards,
Here is a table that should help
\begin{array}{|c|c|c|} \hline P & Q & P\Rightarrow Q \\ \hline T & T & T \\ \hline T & F & F \\ \hline F & T & T \\ \hline F & F & T \\ \hline \end{array}
For the first row, $P$ and $Q$ are both assumed true thus $P \Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $P\Rightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $P\Rightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $P\Rightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.
As an extension to this exercise try this one to solidify your understanding:
\begin{array}{|c|c|c|} \hline P & Q & Q\Rightarrow P \\ \hline T & T & \\ \hline T & F & \\ \hline F & T & \\ \hline F & F & \\ \hline \end{array}
One rather intuitive understanding is that the equivalence $\Leftrightarrow$ can be understood as a conjunction of $\Leftarrow$ and $\Rightarrow$ thus $F\Rightarrow X$ has to be true independent of whether $X$ stands for true or false. As long as the implication does not get violated by a clear example that falsifies it ($T\Rightarrow F$) it is considered to be true.