SOLUTION: Construct a truth table for ~(p<font face = "symbol">Ù</font>q)

Start out with this table:
        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |     |        |
case 2 | T | F |     |        | 
case 3 | F | T |     |        |
case 4 | F | F |     |        |

  

(Rule for "AND", Ù:
Ù requires truth on both sides in
order to be true)

in case 1, p has value T and q has value T, so
Ù has trues on both sides so we put a T under pÙq
in case 1.

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |        |
case 2 | T | F |     |        | 
case 3 | F | T |     |        |
case 4 | F | F |     |        |

in case 2, p has value T and q has value F, so
Ù does NOT have trues on both sides so we put a 
F under pÙq in case 2.

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |        |
case 2 | T | F |  F  |        | 
case 3 | F | T |     |        |
case 4 | F | F |     |        |

in case 3, p has value F and q has value T, so
Ù does NOT have trues on both sides so we put a 
F under pÙq in case 3.

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |        |
case 2 | T | F |  F  |        | 
case 3 | F | T |  F  |        |
case 4 | F | F |     |        |

in case 4, p has value F and q has value F, so
Ù does NOT have trues on both sides so we put a 
F under pÙq in case 4.

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |        |
case 2 | T | F |  F  |        | 
case 3 | F | T |  F  |        |
case 4 | F | F |  F  |        |


Now we fill in the column under ~(pÙq)

(Rule: "NOT" means to give the opposite
value of what follows the symbol "~". That
is, if ~ precedes T, then that gives F and
if ~ precedes F, then that gives T)

Since in case 1, pÙq has value T, then
~(pÙq) will have the opposite value F:

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |   F    |
case 2 | T | F |  F  |        | 
case 3 | F | T |  F  |        |
case 4 | F | F |  F  |        |

Since in case 2, pÙq has value F, then
~(pÙq) will have the opposite value T:

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |   F    |
case 2 | T | F |  F  |   T    | 
case 3 | F | T |  F  |        |
case 4 | F | F |  F  |        |

Since in case 3, pÙq has value F, then
~(pÙq) will have the opposite value T:
        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |   F    |
case 2 | T | F |  F  |   T    | 
case 3 | F | T |  F  |   T    |
case 4 | F | F |  F  |        |

Since in case 4, pÙq has value F, then
~(pÙq) will have the opposite value T:

        _______________________
       | p | q | pÙq | ~(pÙq) | 
case 1 | T | T |  T  |   F    |
case 2 | T | F |  F  |   T    | 
case 3 | F | T |  F  |   T    |
case 4 | F | F |  F  |   T    |

So the truth table for ~(pÙq) is FTTT

Edwin