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 #