Question

How can you use a truth table to prove that `((~p vv q) ^^ p) vv q` is equivalent to `q` ?

Answer 1

See explanation...

Explanation:

In this truth table, `0` represents false, `1` represents true.

`((p,color(blue)(q),~p,~p vv q,(~p vv q) ^^ p,color(green)([(~p vv q) ^^ p] vv q)),(0,color(blue)(0),1,1,0,color(green)(0)),(0,color(blue)(1),0,1,0,color(green)(1)),(1,color(blue)(0),0,0,0,color(green)(0)),(1,color(blue)(1),0,1,1,color(green)(1)))`

The truth values are evaluated for all possible combinations of `p, q` true or false.

Notice that the resulting green column is the same as the blue column.

Answer 2

`[(not p uu q) nn p]uu q = (q nn p) uu q equiv q`

Explanation:

`(not p uu q) nn p = (not p nn p) uu (q nn p)`

but `(not p nn p) = O/` then

`(not p uu q) nn p = (q nn p)`

so

`[(not p uu q) nn p]uu q = (q nn p) uu q equiv q`