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

2 Answers
Jun 20, 2016

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.

Jun 20, 2016

[(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