truth table?

Question

Construct the truth table for the statement
[p^(qv~p)]vr

Answer 1

[p&(qv~p)]v r
1 1 1 1 0 |1| 1
1 1 1 1 0 |1| 0
1 0 0 0 0 |1| 1
1 0 0 0 0 |0| 0
0 0 1 1 1 |1| 1
0 0 1 1 1 |0| 0
0 0 0 1 1 |1| 1
0 0 0 1 1 |0| 0

Answer 2

p q (-q ^ p) -> ~q T T T F F T F F One key element to appreciate approximately conditionals is they're fake in basic terms in one case: T --> F. In all different situations, the conditional is actual. For conditionals, word that as quickly through fact the resultant is actual, the whole conditional is actual. intending to be certain that ~q to be actual, q must be fake. For the two situations q is fake, the conditional would be actual. p q (-q ^ p) -> ~q T T T F . . . . . T . . . . . . F T F F . . . . . T . . . . . . For the different 2 situations, all we ought to do is plug interior the boolean values. If p = T and q = T, then we've (~q ^ p) --> ~q (~T ^ T) --> (~T) (F ^ T) --> F F --> F, that's a real assertion. p q (-q ^ p) -> ~q T T . . . . . T . . . . . . T F . . . . . T . . . . . . F T F F . . . . . T . . . . . . Now, evaluate the case the place p = F and q = T. Then (~q ^ p) --> ~q (~T ^ F) --> ~T (F ^ F) --> F F --> F, that's a real assertion. subsequently: p q (-q ^ p) -> ~q T T . . . . . T . . . . . . T F . . . . . T . . . . . . F T . . . . . T . . . . . . F F . . . . . T . . . . . .

Answer 3

You should start with the distributive rule: http://en.wikipedia.org/wiki/Boolean_algebra#Formal_definition
to simplify the equation, or simply brute force it. With 3 variables, there are only 8 calculations to do:
000
001
010
011
100
101
110
111

Answer 4

I don't understand, what does this have to do with parapsychology?

BB

Answer 5

why are all these in parapsycology???