Question 15316 – DSSSB PGT 2021
March 21, 2024UGC NET CS 2013 Dec-paper-2
March 21, 2024GATE 2021 CS-Set-1
| Question 17 |
Which one of the following choices is correct?
| Both S1and S2 are tautologies. | |
| Neither S1and S2 are tautology. | |
| S1is not a tautology but S2is a tautology. | |
| S1is a tautology but S2is not a tautology. |
A tautology is a formula which is “always true” . That is, it is true for every assignment of truth values to its simple components.
Method 1:
S1: (~p ^ (p Vq)) →q
The implication is false only for T->F condition.
Let’s consider q as false, then
(~p ^ (p Vq)) will be (~p ^ (p VF)) = (~p ^ (p)) =F.
It is always F->F which is true for implication. So there are no cases that return false, thus its always True i.e. its Tautology.
S2:
q->(~p (p Vq))
The false case for implication occurs at T->F case.
Let q=T then (~p (p Vq)) = (~p (p VT))= ~p. (It can be false for p=T).
So there is a case which yields T->F = F. Thus its not Valid or Not a Tautology.
Method 2:
A tautology is a formula which is “always true” . That is, it is true for every assignment of truth values to its simple components.
Method 1:
S1: (~p ^ (p Vq)) →q
The implication is false only for T->F condition.
Let’s consider q as false, then
(~p ^ (p Vq)) will be (~p ^ (p VF)) = (~p ^ (p)) =F.
It is always F->F which is true for implication. So there are no cases that return false, thus its always True i.e. its Tautology.
S2:
q->(~p (p Vq))
The false case for implication occurs at T->F case.
Let q=T then (~p (p Vq)) = (~p (p VT))= ~p. (It can be false for p=T).
So there is a case which yields T->F = F. Thus its not Valid or Not a Tautology.
Method 2: