SOLUTION: Given �(p∧q), prove (�p∨�q).

Question 664426: Given �(p∧q), prove (�p∨�q).
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

�(p∧q), prove (�p∨�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∧y is: If x and y both have
   T's, x∧y has T, otherwise x∧y has F.  
3. Disjunction: Rule for x∨y is: If x and y both have 
   F's, x∨y has F, otherwise x∨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∧q | �(p∧q) | �p∨�q | �(p∧q) <-> (�p∨�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∧q is because there is a T under both p and q,
    rule 2 above.
The F under �(p∧q) is because there is a T under p∧q,
    rule 1 above.
The F under �p∨�q is because �p and �q both have F's
    rule 3 above.
The T under �(p∧q) <-> (�p∨�q) is because both �(p∧q) and (�p∨�q)
    have the same truth value F, rule 4

Now we fil in the next row:

p | q | �p | �q | p∧q | �(p∧q) | �p∨�q | �(p∧q) <-> (�p∨�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∧q is because p and q don't both have T's, rule 2 
above.
The T under �(p∧q) is because there is a F under p∧q,
    rule 1 above.
The T under �p∨�q is because �p and �q don't both have F's
    rule 3 above.
The T under �(p∧q) <-> (�p∨�q) is because both �(p∧q) and (�p∨�q)
    have the same truth value T, rule 4

Now we fil in the third row:

p | q | �p | �q | p∧q | �(p∧q) | �p∨�q | �(p∧q) <-> (�p∨�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∧q is because p and q don't both have T's, rule 2 
above.
The T under �(p∧q) is because there is a F under p∧q,
    rule 1 above.
The T under �p∨�q is because �p and �q don't both have F's
    rule 3 above.
The T under �(p∧q) <-> (�p∨�q) is because both �(p∧q) and (�p∨�q)
    have the same truth value T, rule 4

Now we fill in the fourth row:

p | q | �p | �q | p∧q | �(p∧q) | �p∨�q | �(p∧q) <-> (�p∨�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∧q is because p and q don't both have T's, rule 2 
above.
The T under �(p∧q) is because there is a F under p∧q,
    rule 1 above.
The T under �p∨�q is because �p and �q don't both have F's
    rule 3 above.
The T under �(p∧q) <-> (�p∨�q) is because both �(p∧q) and (�p∨�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∧q) <-> (�p∨�q) is
true and therefore �(p∧q) and (�p∨�q) are equivalent
because the biconditional is a tautology.

Edwin

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SOLUTION: Given �(p&#8743;q), prove (�p&#8744;�q).

SOLUTION: Given �(p∧q), prove (�p∨�q).