Posted - 07/14/2011 : 16:36:23 Construct a truth table for the statement.

(p ↔ q) → p

I got this:

p. q. p<->q T. T. T T. F. F F. T. F F. F. T

But I don't believe it's correct....please help!

2 L A T E S T R E P L I E S (Newest First)

the_hill1962

Posted - 07/19/2011 : 15:46:40 Yes, your table of p <--> q is correct. Now, the original problem was to put that into an implication with -->p After you do the interactive lesson that Admin referenced, you will see (or maybe already knew) that p-->q has a table of p q ..p-->q T T .. T T 0 .. 0 0 T .. T 0 0 .. T So using your table of p q ..p<->q T T .. T T F .. F F T .. F F F .. T in the problem (p ↔ q) → p makes it look like the following: p q ..(p<->q) T T ... T T F ... F F T ... F F F ... T and then taking the list for (p<->q) with what "p" was gives: (p<->q) --> p . T ...... T . F ...... T . F ...... F . T ...... F So, all you have to to is list the values for ?#? below. T-->T is ?1? F-->T is ?2? F-->F is ?3? T-->F is ?4? and that will be the values to finish: p q ..(p ↔ q) → p T T ..... ?1? T 0 ..... ?2? 0 T ..... ?3? 0 0 ..... ?4?