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gurly691
Average Member

USA
10 Posts

 Posted - 07/14/2011 :  16:36:23 Construct a truth table for the statement.(p ↔ q) → pI got this:p. q. p<->qT. T. TT. F. FF. T. FF. F. TBut I don't believe it's correct....please help!

USA
635 Posts

 Posted - 07/14/2011 :  16:43:28 quote:Originally posted by gurly691Construct a truth table for the statement.(p ¡ê q) ¡æ pI got this:p. q. p<->qT. T. TT. F. FF. T. FF. F. TBut I don't believe it's correct....please help!See our interactive lessons on Logic.Gisele

the_hill1962

USA
1461 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 -->pAfter you do the interactive lesson that Admin referenced, you will see (or maybe already knew) that p-->q has a table ofp q ..p-->qT T .. TT 0 .. 00 T .. T0 0 .. TSo using your table ofp q ..p<->qT T .. TT F .. FF T .. FF F .. Tin the problem (p ↔ q) → pmakes it look like the following:p q ..(p<->q)T T ... TT F ... FF T ... FF F ... Tand then taking the list for (p<->q) with what "p" was gives:(p<->q) --> p . T ...... T . F ...... T . F ...... F . T ...... FSo, 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) → pT T ..... ?1?T 0 ..... ?2?0 T ..... ?3?0 0 ..... ?4?
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