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| gurly691 |
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?
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| Admin |
Posted - 07/14/2011 : 16:43:28
quote:
Originally posted by gurly691
Construct a truth table for the statement.
(p ¡ê q) ¡æ 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!
See our interactive lessons on Logic.
Gisele |
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