Construct a truth table for (~q ^ p) → ~q?

Question

Please work out, so I can understand it. Thanks

Answer 1

Hi,

Remember:
1) and (conjunction) is true only when both p and q are true.
2) The conditional --> is false only when the antecedant ((~q^p) in this case) is true and the consequent, ~q is false.
So, let's make our table. The answer will be under the conditional sign, -->



p...q...~q..(~q^p)..-->.~q
T...T....F.....F.......T....F
T...F...T......T......T....T
F.. T...F......F......T....F
F...F..T...... F.....T....T


So, the answer is all true.

I hope this table comes out okay. Anyone know an easy way to make a truth table on this system?
Hope this helps.

FE

Answer 2

p | q | ~q | (~q ^ p) | (~q ^ p) → ~q
T...T...F.....F&T=F......... F→T=T
T...F...T.....T&T=T......... T→T=T
F...T...F.....F&F=F......... F→F=T
F...F...T.....T&F=F.......... F→T=T

This is a tautology, ie it is always true.