Question 664426
<pre>
¬(p&#8743;q), prove (¬p&#8744;¬q).

This is done by a truth table.

1. Negation:  Rule for ¬x is: If x has a T, then ¬x has
   an F and vice-versa.
2. Conjunction: Rule for x&#8743;y is: If x and y both have
   T's, x&#8743;y has T, otherwise x&#8743;y has F.  
3. Disjunction: Rule for x&#8744;y is: If x and y both have 
   F's, x&#8744;y has F, otherwise x&#8744;y has T.
4. Biconditional:  Rule for x<->y is: If both x and y 
   have the same truth value, x<->y has T, otherwise F. 

You need headings as listed below:   

p | q | ¬p | ¬q | p&#8743;q | ¬(p&#8743;q) | ¬p&#8744;¬q | ¬(p&#8743;q) <-> (¬p&#8744;¬q) |
---------------------------------------------------------------
T | T |  F |  F |  T  |    F    |   F   |         T           |
T | F |    |    |     |         |       |                     |
F | T |    |    |     |         |       |                     |
F | F |    |    |     |         |       |                     |

The F under ¬p is because there is a T under p, rule 1 above.
The F under ¬q is because there is a T under q, rule 1 above.
The T under p&#8743;q is because there is a T under both p and q,
    rule 2 above.
The F under ¬(p&#8743;q) is because there is a T under p&#8743;q,
    rule 1 above.
The F under ¬p&#8744;¬q is because ¬p and ¬q both have F's
    rule 3 above.
The T under ¬(p&#8743;q) <-> (¬p&#8744;¬q) is because both ¬(p&#8743;q) and (¬p&#8744;¬q)
    have the same truth value F, rule 4

Now we fil in the next row:

p | q | ¬p | ¬q | p&#8743;q | ¬(p&#8743;q) | ¬p&#8744;¬q | ¬(p&#8743;q) <-> (¬p&#8744;¬q) |
---------------------------------------------------------------
T | T |  F |  F |  T  |    F    |   F   |         T           |
T | F |  F |  T |  F  |    T    |   T   |         T           |
F | T |    |    |     |         |       |                     |
F | F |    |    |     |         |       |                     |

The F under ¬p is because there is a T under p, rule 1 above.
The T under ¬q is because there is a F under q, rule 1 above.
The F under p&#8743;q is because p and q don't both have T's, rule 2 
above.
The T under ¬(p&#8743;q) is because there is a F under p&#8743;q,
    rule 1 above.
The T under ¬p&#8744;¬q is because ¬p and ¬q don't both have F's
    rule 3 above.
The T under ¬(p&#8743;q) <-> (¬p&#8744;¬q) is because both ¬(p&#8743;q) and (¬p&#8744;¬q)
    have the same truth value T, rule 4

Now we fil in the third row:

p | q | ¬p | ¬q | p&#8743;q | ¬(p&#8743;q) | ¬p&#8744;¬q | ¬(p&#8743;q) <-> (¬p&#8744;¬q) |
---------------------------------------------------------------
T | T |  F |  F |  T  |    F    |   F   |         T           |
T | F |  F |  T |  F  |    T    |   T   |         T           |
F | T |  T |  F |  F  |    T    |   T   |         T           |
F | F |    |    |     |         |       |                     |

The T under ¬p is because there is a F under p, rule 1 above.
The F under ¬q is because there is a T under q, rule 1 above.
The F under p&#8743;q is because p and q don't both have T's, rule 2 
above.
The T under ¬(p&#8743;q) is because there is a F under p&#8743;q,
    rule 1 above.
The T under ¬p&#8744;¬q is because ¬p and ¬q don't both have F's
    rule 3 above.
The T under ¬(p&#8743;q) <-> (¬p&#8744;¬q) is because both ¬(p&#8743;q) and (¬p&#8744;¬q)
    have the same truth value T, rule 4

Now we fill in the fourth row:

p | q | ¬p | ¬q | p&#8743;q | ¬(p&#8743;q) | ¬p&#8744;¬q | ¬(p&#8743;q) <-> (¬p&#8744;¬q) |
---------------------------------------------------------------
T | T |  F |  F |  T  |    F    |   F   |         T           |
T | F |  F |  T |  F  |    T    |   T   |         T           |
F | T |  T |  F |  F  |    T    |   T   |         T           |
F | F |  T |  T |  F  |    T    |   T   |         T           |

The T under ¬p is because there is a F under p, rule 1 above.
The T under ¬q is because there is a F under q, rule 1 above.
The F under p&#8743;q is because p and q don't both have T's, rule 2 
above.
The T under ¬(p&#8743;q) is because there is a F under p&#8743;q,
    rule 1 above.
The T under ¬p&#8744;¬q is because ¬p and ¬q don't both have F's
    rule 3 above.
The T under ¬(p&#8743;q) <-> (¬p&#8744;¬q) is because both ¬(p&#8743;q) and (¬p&#8744;¬q)
    have the same truth value T, rule 4.

Now since there are only T's in the last column, that proves that
in every case, the biconditional ¬(p&#8743;q) <-> (¬p&#8744;¬q) is
true and therefore ¬(p&#8743;q) and (¬p&#8744;¬q) are equivalent
because the biconditional is a tautology.

Edwin</pre>