Complete the following table, filling
in just for the main connectives.
In the answer key I have indicated the values that combine for the main
connective in red and the value of the main connective itself in blue.
P
|
Q
|
R
|
P->(Q
& R)
|
P->(Q
v R)
|
(P
v Q) -> R |
(P
&~Q) -> ~R
|
~P
& (Q <-> R)
|
(P
v~Q) v (R & ~P)
|
1
|
1
|
1
|
1 1 1
|
1 1 1
|
1 1
1 |
0 1 0
|
0 0
1
|
1 1
0
|
1
|
1
|
0
|
1 0
0
|
1 1 1
|
1 0
0
|
0 1 1
|
0 0
0
|
1 1
0
|
1
|
0
|
1
|
1 0
0
|
1 1 1
|
1 1 1
|
1 0 0
|
0 0
0
|
1 1
0
|
1
|
0
|
0
|
1 0
0
|
1 0 0
|
1 0 0
|
1 1 1
|
0 0
1
|
1 1
0
|
0
|
1
|
1
|
0 1
1
|
0 1 1
|
1 1
1
|
0 1 0
|
1 1
1
|
0 1
1
|
0
|
1
|
0
|
0 1
0
|
0 1
1
|
1 0
0
|
0 1 1
|
1 0
0
|
0 0 0
|
0
|
0
|
1
|
0 1 0
|
0 1
1
|
0 1
1
|
0 1 0
|
1 0
0
|
1 1 1
|
0
|
0
|
0
|
0 1
0
|
0 1 0
|
0 1
0
|
0 1 1
|
1 1
1
|
1 1
0
|
Do
note that the most common errors involve the use of the arrow. It
may seem odd to think that if the antecedent of a conditional is false,
then the entire statement has to be seen as true. The point,
however, is that only situation that permits us to see a conditional as
false is to meet the suffcient condition expressed in the antecedent
and not have the result expressed in the consequent. Imagine
saying that if Alice studies she will pass. When Alice fails to
study we have no way of saying that the relationship was false, no
matter whether she manages to pass or not. Since all expressions
must be seen as either true or false, not being clearly false means we
accept it as true by default. (This is one reason I prefer to use
1 and 0 instead of T and F: we do not need to think of what it means to
say something is "really" true.)
