# Make the truth table of the proposition ¬q→[(pΛq)V~p]?

Apr 1, 2018

See below.

#### Explanation:

Given: not p -> [(p ^^ q)vv ~p]

Logic operators:" not p:" not p, ~ p; " and:" ^^ ; or:vv

Logic Tables, negation:
ul(|" "p|" "q|" "~p|" "~q|)
$\text{ "T|" "T|" "F|" } F |$
$\text{ "T|" "F|" "F|" } T |$
$\text{ "F|" "T|" "T|" } F |$
$\text{ "F|" "F|" "T|" } T |$

Logic Tables, and & or:
$\underline{| \text{ "p|" "q|" "p^^q" "|" "qvvq" } |}$
$| \text{ "T|" "T|" "T" "|" "T" } |$
$| \text{ "T|" "F|" "F" "|" "T" } |$
$| \text{ "F|" "T|" "F" "|" "T" } |$
$| \text{ "F|" "F|" "F" "|" "F" } |$

Logic Tables, if then:
$\underline{| \text{ "p|" "q|" "p->q" } |}$
$| \text{ "T|" "T|" "T" } |$
$| \text{ "T|" "F|" "F" } |$
$| \text{ "F|" "T|" "T" } |$
$| \text{ "F|" "F|" "T" } |$

Given Logic proposition part 1:
ul(|" "p^^q" "|" "~p" "|" "(p^^q)vv~p|)
$| \text{ "T" "|" " F " "|" "T" } |$
$| \text{ "F" "|" " F " "|" "F" } |$
$| \text{ "F" "|" " T " "|" "T" } |$
$| \text{ "F" "|" " T " "|" "T" } |$

Given Logic proposition part 2:
ul(|" "~q" "|" "(p^^q)vv~p|" "~q->(p^^q)vv~p|)
$| \text{ " F " "|" "T" "|" "T" } |$
$| \text{ " T " "|" "F" "|" "F" } |$
$| \text{ " F " "|" "T" "|" "T" } |$
$| \text{ " T " "|" "T" "|" "T" } |$