Question

conditional

Answer 1

\[p \rightarrow q = ?\]
\[q \rightarrow p\]
\[\neg p \rightarrow \neg q\]
\[\neg q \rightarrow \neg p\]

Answer 2

c

Answer 3

am I right or am I right? laughing out loud

Answer 4

last one

Answer 5

is there a way to prove this without a truth table?

Answer 6

I would rephrase p->q as (p AND q') OR p'. Then, boolean algebra may be used to arrive at q'->p'.

Answer 7

But, in this case, a truth table is a perfectly legitimate (and easiest) proof since we are only talking about a boolean operation on two variables.

Answer 8

\[\begin{split}
p\to q &\iff \lnot ~p\lor q \\
&\iff \lnot ~\lnot ~p\land \lnot~q\\
&\iff \lnot ~q\land p\\
&\iff \lnot ~\lnot ~q\lor \lnot ~p \\
&\iff \lnot ~q\to \lnot ~p \\
\end{split}\]

Answer 9

Yeah, wio's simplification of p->q was definitely preferrable, although both work.

Answer 10

i was just doing that

Answer 11

i want to medal you again, man laughing out loud