edited by
447 views

2 Answers

Best answer
3 votes
3 votes
p q p → q p ∧ ( p → q ) (p ∧ ( p → q )) → q
0 0 1 0 1
0 1 1 0 1
1 0 0 0 1
1 1 1 1 1

Since , all are 1, hence tautology

selected by
2 votes
2 votes

(p ∧ ( p → q )) → q

≡(p ∧ ( ~ p ∨ q )) → q

≡~(p ∧ ( ~ p ∨ q ))  ∨ q

≡(~p ∨ (  p ∧ ~q ))  ∨ q

≡(~p ∨ p ) ∧ ( ~p ∨ ~q ) ∨ q

≡( ~p ∨ ~q ) ∨ q

≡~p ∨ (~q ∨ q)

≡~p ∨ T

≡T

Hence,(p ∧ ( p → q )) → q is Tautology. 

Related questions

0 votes
0 votes
1 answer
2
Shawn Frost asked May 20, 2019
438 views
I Have doubt about the language. Is it asking about the sum of elements if we make the GBL set for the given lattice .
0 votes
0 votes
0 answers
3
Ritabrata Dey asked Mar 21, 2019
323 views
I doubtWhether (a,b)R(b,c) is symmetric or antisymmetric or reflexive relationAnd how to approach this type of sums?
1 votes
1 votes
1 answer
4
mathematics asked Oct 19, 2017
311 views
What is the upper bound for the Chromatic Number given by Brooks' theorem for the Petersen graph? (a) 2 (b) 3 (c) 4 (d) None of the above