1.2 12D In Progress

1 [(p v q) ^ (p => r) ^ (q => r)] => r
2 [(p v q) ^ (~p v r) ^ (~q v r)] => r by OULE
3 ~[(p v q) ^ (~p v r) ^ (~q v r)] v r by OULE
[~(p v q) ^ ~(~p v r) v ~(~q v r)] v r by De Morgan's Law (I think this line should replace step 4.)
[(~p ^ ~q) ^ (~p ^ ~r) v (q ^ ~r)] v r by De Morgan's Law and Double Negation
[(~p ^ ~p) ^ (~q ^ ~r) v (q ^ ~r)] v r by Commutative and Association Laws
[~p ^ (~q ^ ~r) v (q ^ ~r)] v r by Idempotent Law
[~p ^ ((~q ^ ~r) v q) ^ ((~q ^ ~r) v ~r))] v r by Distributive Law
[~p ^ ((~q v q) ^ (~r v q)) ^ ((~q v ~r) ^ (~r v ~r))] v r by Distributive Law
[~p ^ ((T) ^ (~r v q)) ^ ((~q v ~r) ^ ~r)] v r by Commutative and Negation Law
[~p ^ (~r v q) ^ ((~q v ~r) ^ ~r)] v r by Identity Law
[((~p ^ ~r) v (~p ^ q)) ^ ((~q v ~r) ^ ~r)] v r by Distributive Law
(r v ((~p ^ ~r) v (~p ^ q))) ^ (r v ((~q v ~r) ^ ~r)) by Commutative and Distributive Laws (Review this line.)
(r v (~p ^ ~r))
4 [(~p v ~q) v (p v ~r) v (q v ~r)] v r by DeMorgans
5 [(~p v p) v (~q v ~r) v (q v ~r)] v r by Associative
6 [(T) v (~q v ~r) v (q v ~r)] v r by Domination
7 [T v (q v ~r)] v r by Domination
8 T v r = T by Domination

Section 1.2 Problem 12 b (my solution)

Prove [(p=>q) ^ (q=>r)] => (p=>r) is a Tautology

Proof: " "

1 = [(~pvq) ^ (~qvr)] => (~pvr) by OULE p=>q = ~pvr
2 =~[(~pvq) ^ (~qvr)} v (~pvr) by OULE " "
3 =[~(~pvq) v ~(~qvr)] v (~pvr) by De Morgan's Law
4 =[(~~p^~q) v (~~q^~r)] v (~pvr) by De Morgan's Law
5 =[(p^~q) v (q^~r)] v (~pvr) by Double Negation Law
6 =[((p^~q) v q ^ ((p^~q) v ~r)] v (~pvr) by Distributive Law
7 =[((pvq) ^ (~qvq)) ^ ((p^~q) v ~r))] v (~pvr) by Distributive Law
8 =[((pvq) ^ T) ^ ((p^~q) v ~r)] v (~pvr) by Negation law
9 =[(pvq) ^ ((p^~q) v ~r)] v (~pvr) by Identity law
10 =(~pvr) v [(pvq) ^ ((p^~q) v ~r)] by Commutative Law
11 =[(~pvr) v (pvq)[ ^ [(~pvr) v ((p^~q) v ~r] by Distributive Law
12 =[(~pvp) v (rvq)] ^ [(rv~r) v (~pv(p^~q))] by Commutative Law
13 =[T v (rvq)] ^ [Tv(~pv(p^~q))] by Negation Law
14 =T^T by Domination Law
15 =T by Idempotent Law or Definition of Conjunction

Section 1.2 Problem 12 b

Prove that [(p -> q) ^ (q -> r)] -> (p -> r) is a tautology.

1 = ~[(~p v q) ^ (~q v r)] v (~p v r) by OULE - (p => q) = ~p v q
2 = [~(~p v q) v ~(~q v r)] v (~p v r) by DeMorgans
3 = [(p ^ ~q) v (q ^ ~r)] v (~p v r) by deMorgans and Double Negation
4 = [((p ^ ~q) v q) ^ ((p ^ ~q) v ~r)] v (~p v r) by commutative and distributive
5 = [((q v p) ^ (q v ~q)) ^ ((p ^ ~q) v ~r)] v (~p v r) by Negation and Identity
6 = [(q v p) ^ ((p ^ ~q) v ~r)] v (~p v r) by commutative and distributive
7 = ((~p v r) v (q v p)) ^ ((~p v r) v ((p ^ ~q) v ~r)) by commutative and associative
8 = (p v ~p) v (q v r) ^ ((r v ~r) v (~p v (p ^ ~q))) by Negation
9 = (T v (q v r)) ^ (T v (~p v (p ^ ~q))) by commutative and domination
10 = T ^ T = T

(taken from another student in class)