Kenneth Rosen Edition 6th Exercise 1.3 Question 29 (Page No. 48)

Question

Express each of these statements using logical operators,
predicates, and quantifiers.
a) Some propositions are tautologies.
b) The negation of a contradiction is a tautology.
c) The disjunction of two contingencies can be a tautol-
ogy.
d) The conjunction of two tautologies is a tautology

https://math.stackexchange.com/questions/219207/predicates-quantifiers

I read above but still  confuse....can somebody explain statement c) and d) in more detail

Answer 1

Let X and Y are prepositions.

Tautology means Every row of X is 1, if X is a Tautology represent by T(X) ===> ~T(X) represent X is not a Tautology but T(~X) represents '~X' is a Tautology.

Contradiction means Every row of X is 0, if X is a Contradiction represent by C(X)  ===> ~C(X) represent X is not a Contradiction but C(~X) represents '~X' is a Contradiction.

Contingency means Preposition is neither Tautology nor Contradiction. ===> ~T(X) ^ ~C(X)

 

C) The disjunction of two contingencies can be a tautology. 

            note that disjunction of two contingencies are not always a tautology it may be a contradiction or contingency.

which means there exist two contingencies which disjunction is tautology ===> X is a Contingency and Y is Contingency but their disjunction is a Tautology

       represent two different contingencies ( ~T(X) ^ ~C(X) ) and  ( ~T(Y) ^ ~C(Y) )  and their disjunction is Tautology T( X V Y )

          ∴ ∃X ∃Y (           ( ~T(X) ^ ~C(X) )  ^  ( ~T(Y) ^ ~C(Y) )  ^      T( X V Y )       )

 

but you may confused with  ∃X ∃Y (       (    ( ~T(X) ^ ~C(X) )  ^  ( ~T(Y) ^ ~C(Y) )  ) ----------->     T( X V Y )       )

 the above formula says that every disjunction of two contingencies can be a tautology

 

D) The conjunction of two tautologies is a tautology

              given statement is conjunction of two tautologies is a tautology always.

represent two different tautologies T(X)  and  T(Y)   and their conjunction is Tautology T( X ^ Y )

          ∴ ∀X ∀Y (           (  T(X)   ^  T(Y)  )  ------------->     T( X ^ Y )       )

Comments

@Shaik Masthan thanks for answer. i was having doubt does i should use negation or conjunction but with this line " disjunction of two contingencies are not always a tautology it may be a contradiction or contingency." it is clear. Thanks.

I have same doubt with 

Q a) When there is less than 30 megabytes free on the hard
disk, a warning message is sent to all users.

Q b) Whenever there is an active alert, all queued messages
are transmitted.

for all this i am confuse does i have to use ----> or  ^ .

lets take first question 

When there is less than 30 megabytes free on the hard
disk, a warning message is sent to all users.

if i use implication as follows 

∃x P(x) ---> ∀y Q(y)  ------does it is correct ?

but ∃x P(x) false still entire statement is true.

It mean even we have more than 30 megabyte free on the hrd disk  warning message will be sent to all user.

answer for other , but i have same doubt.

Q b) ∃x (A(x) ---> ∀y S(x,y)) ---------so even if there is no active alert , message still sent to all user.

Can you Please explain this. Thanks, you can  edit your answer :)  

 

---

Those 2 questions are similar...

Whenever you use ∃ you have to use ^ connector

Whenever you use ∀ you have to use -> connector

Therefore answer for

first question is ∃x P(x) ^ ∀y Q(y)

Second question is ∃x (A(x) ^ ∀y S(x,y))