SOLUTION: Construct a truth table for p -> (~q ^ p)

Question 310536: Construct a truth table for p -> (~q ^ p)
Answer by Edwin McCravy(20055) (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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