Question

show that \[\neg(p\leftrightarrow q) \equiv p \leftrightarrow \neg q\]

i got to a part that says \[(p\vee q) \wedge (\neg q \vee \neg p)\]and i can smell i'm close. so what's next?

Answer 1

aaarrrrggggggghhhhhhhhhh

Answer 2

now i have \[(\neg p \rightarrow q) \wedge (q \rightarrow \neg p)\]

Answer 3

so i assume this means \[\neg p \leftrightarrow q\]yes?

Answer 4

sadly....that's not the original...

Answer 5

left side is
\[\lnot [(p\land q)\lor (\lnot p \land \lnot q)]\]

Answer 6

hmm

Answer 7

right side is
\[(p\land \lnot q)\lor (\lnot p \land q)\]

Answer 8

you have to work towards something right? i mean you have to show "this mess is equivalent to that other mess"

Answer 9

i just have to solve one side and make it look like the original

Answer 10

rigth, but we need to know what both sides look like with and and or statements to do that

Answer 11

i think in the text i have it simply states this as a "logical equivalence of biconditonal statements" but if you are not going to show it with truth tables, you have to arrive at it somehow

Answer 12

ahh nevermind. got it

Answer 13

okay!