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Question 310536: Construct a truth table for p -> (~q ^ p)
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
Learn the rules:
~ always changes T to F and F to T, that is it always gives the opposite.
V is usually T. The only times V is F is when there is an F on both sides of V
^ is usually F. The only times ^ is T is when there is a T on both sides of ^
-> is usually T. The only times -> is F is when there is a T on the left of -> and a F on the right of ->
That is, we learn the exceptional cases for V, ^ and ->. They are
F V F becomes F, other cases T
T ^ T becomes T, other cases F
T -> F becomes F, other cases T
p -> (~q ^ p)
Start with this, so that you have all the "pieces", that is,
You can build ~q from q.
You can build (~q ^ p) from ~q and p.
You can build p -> (~q ^ p) from p and (~q ^ p)
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
| | | | |
| | | | |
| | | | |
| | | | |
Put TTFF under p and TFTF under q
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | | | |
T | F | | | |
F | T | | | |
F | F | | | |
Now since q has TFTF under it, ~q must have the opposites,
FTFT under it:
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | | |
T | F | T | | |
F | T | F | | |
F | F | T | | |
To fill in the next column, we use the rule for ^ which is
Put an F unless there is a T
on the left and a T on the right. Only in that one case do we put T
In the top row ~q is F and p is T, so we must put F, because there is
not a T on both sides of ^. This is not the exceptional case for ^.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | |
T | F | T | | |
F | T | F | | |
F | F | T | | |
In the second row ~q is T and p is T, so we must put T, because there is
a T on both sides of ^ and that is the one case of ^ when we must put a T.
This is the exceptional case for ^.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | |
T | F | T | T | |
F | T | F | | |
F | F | T | | |
In the third row ~q is F and p is F, so we must put F, because there is
not a T on both sides of ^. This is not the exceptional case for ^.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | |
T | F | T | T | |
F | T | F | F | |
F | F | T | | |
In the bottom row ~q is T and p is F, so we must put F, because there is
not a T on both sides of ^. This is not the exceptional case for ^.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | |
T | F | T | T | |
F | T | F | F | |
F | F | T | F | |
To fill in the last column, we use the rule for -> which is
Put a T unless there is a T
on the left and a F on the right. Only in that one case do we put F
On the top line we put F because there is a T under p and an F under (~q ^ p),
and that is the one case when we must put F under ->,
This IS the exceptional case for ->.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | F |
T | F | T | T | |
F | T | F | F | |
F | F | T | F | |
On the second line we put T because there is a T under p and a T under
(~q ^ p). This is not the exceptional case for ->.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | F |
T | F | T | T | T |
F | T | F | F | |
F | F | T | F | |
On the third line we put T because there is an F under p and
an F under (~q ^ p). This is not the exceptional case for ->.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | F |
T | F | T | T | T |
F | T | F | F | T |
F | F | T | F | |
Finally, on the fourth line we put T because there is an F under p and
an F under (~q ^ p). This is not the exceptional case for ->.
p | q | ~q | (~q ^ p) | p -> (~q ^ p) |
--|---|----|----------|---------------|
T | T | F | F | F |
T | F | T | T | T |
F | T | F | F | T |
F | F | T | F | T |
Edwin
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