Question

show that ¬(p↔q)and¬p↔q are logically equivalent

Answer 1

\[\begin{array}{c|c|c|c}p&q&p\leftrightarrow q & \lnot(p\leftrightarrow q) \\\hline T &T &T & F\\ T & F & F & T \\ F&T & F & T \\ F&F&T&F\end{array}\]\[\begin{array}{c|c|c|c}p&q&\lnot p & \lnot p\leftrightarrow q \\\hline T &T &F & F\\ T & F & F & T \\ F&T & T & T \\ F&F&T&F\end{array}\]

Answer 2

(\(\lnot (p\leftrightarrow q\)\) and \(\lnot p\leftrightarrow q\) are logically equivalent because the two last columns in those truth tables are the same when the first two columns were kept the same as well.)

Answer 3

thanks
but I have to do it with laws
not truth tables