Express the propositions and using only and
where and are propositions, it's defined by the following truth table:
Let be propositions such that is false. Determine the value of truth of:
Let and be propositions such that is false. Give the value of truth of the following propositions (justify your answer):
I'm learning these stuff, so if you can help me, it'd be great. Thanks
what does veebar mean? Is that "exclusive or" ?
Originally Posted by Krizalid
I'm not sure i understood well these notations o.O
What if you make a table of truth for A => B, which is ¬P v Q ?
A=1, B=1, (A => B) = 1
A=0, (A => B) = 1
A=1, B=0, (A => B) = 0
For the first one, find the values of p1 to p4 such as the proposition is false. Then you can determine the table of truth of i. ii. and iii. according to the values of p1 to p4...
is ; hence is .
is so that would be .
I don't know of any direct method; I used some good guessing . . .
For the first one, we know the pattern: .
Now we must come up with using only and
With a little testing I found that has
. . . .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the second one: .
I found that . .produces the same pattern: .
Originally Posted by Plato
The truth table is exactly the same as the logic of a NOR gate in electronics.
I have explanations for the third problem,
. . but I don't know if they are acceptable for your course.
I also used the alternate defintion of implication:
An implication has this truth table: .
. . An implication is false if the first part is true and the second part is false.
. . An implication is true if the first part is false.
(i) If is true, then is true.
. . But is equivalent to:
. . Therefore: . is true.
(ii) We are given: .
. . And we are told that is false.
. . Since the first part is false, the implication is true.