the_hill1962
Advanced Member
USA
1468 Posts 
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?


