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asked in Mathematical Logic by Boss (7.4k points)
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All are invalid
Please explain why
@arjun sir,pls help in this one.
sorry, I dont want to boil my brain solving such questions. For mathematical logic there are numerous previous year questions and also many PDFs are given in gatecse. When all those are there I do not want to waste time decoding such poorly formatted questions.
i understand arjun sir,but if u colud just tell which option is correct then that would have helped..!

thanks anyway :-)
^That means you did not understand :(

1 Answer

+1 vote
Statement 1)

Let the domain be set of real numbers.

B: true

A: number is integer

LHS = false and RHS= true

So, fale double implies true is false.

 

Statement 2)

Let the domain be set of real numbers.

B: false

A: number is integer

LHS = true and RHS= false

So, true double implies false is false.
answered by Veteran (18k points)
in first option,B is free from any variable..so whether it is E(x) oR V(x)..it should not effect the meanig of statement..so i think statement 1st should be valid as B is not bound.so whther it is true for one value or all values of x,it does not matter.

statement 2:LHS means that if A is true for some x then B is true,which means ,take a set x=(3,6,2,9,1) and A(x)=div by 3,B =sun shines in the noon.now A hold true for some values in x viz 3,6,9 .so from statement B is true

                  RHS means if for every x ,A is true then B is true...lets take x=3 ,A(x) is true,so B is true,

for x=6,A is true,hence B is true.same for x=9.

so,i think statment 2 id also valid.this is also because here B is again not bound.it is a free variable

-----------------------------------------------------------------------------

now third statement says:

LHS -if for some x ,A(x) is true then B(x) is also true.

RHS says,if A(x) is true for all X the B(x) is true for some x.

LHS is different from RHS here.

so i guess first two statements are coorect
in your proof for statement 1,how is LHS false and RHS true.

A will be always true according to your set as every integer is a real number
LHS is  saying:

for all x,  B -> A(x)

Now, if I take x=1.5, above statement will be false.

Similarly, for RHS
LHS does not mean that if B is true for all the values of x then A has to be true for "all the values of x" because for all quantifier is not with A..
how to solve sch type of question then..i am really confused
and can you tell any one example in which statemnt 2 dissatifies..?

as in your example,you have taken B to be false then how can LHS Be false?

LHS-for some x if A(x) is true then B is true...obviously for some x,A(x) is true,so true -> false is false.hence,LHS is false

RHS ---for each x,if A(x) is true then B is true,then this false

and false <=> is false is true
@Akriti. Lets see for 1st example.

Since 'x' isnt associated with B, you can any value for B. I think its like universal truth which doesnt depend upon anything.

Now, for 2nd example.

I want to prove that this double implication is false. I want to find such an instance. Thats what I am doing.

Since, the quantifier is 'there exists', LHS must be true for atleast one instance and such instance exists.
by the way, I am seeing such an example for 1st time where B is used without the domain variable :P
sushant,

in first statement,universal quantifier's scope is not covering A.it is  [V(x)B -> A(x)]---here you can see that V(x) is only with B not A,though B is free from x.so,LHS is just saying that if B holds for all the values then A also holdsit does not mean that A should hold for all the values.

and in second statement -

LHS is saying if (A holds for ssome values )then (B also holds),here you have taken B as False,hence,LHS becomes false.how is it TRue..??as A is true for some values atleast as some real number are integer as well.so,T -> F is False.

and in RHS,it is saying for all x,if A is true then B is true,which is again False .

so false <=> false is TRUE..right??
@Akriti. If scope was limited to B, then from where would you get 'x' for A? :)
and even if in your second statement,you will change B to true

then LHS will b true and RHS will also be true
so,do you mean V(x)B -> A(x) and V(x) ( B -> A(x) ) are same..??

if it is so,then yes this is not correct.

can you tell about statment 2..??
@Akriti. You need to find example by trial and error to disprove the things.  Even I am slow in doing so.

Could you try some examples like the one in 2016 exam I think ?
Just notice the difference here:

Vx B -> A(x)

 and

Vx B  -> Vx A(x)

did you get it?
@sushant,yes,i have gone through the above link and see in that one 3rd statment is true.

meanwhile i understood that statement 1 is incorrect but

for statement 2,i was thinking that as B is already true then whether it is for some A to be true or for all A to be true,it is true afterall as it is universal.i guess i am wrong only.:-(

THanks for helping out!

3rd statment is wrong na??
Lhs

= $\exists$x( A(x) -> B(x) )

= $\exists$x( ~A(x) V B(x) )

= $\exists$x ~A(x)  V  $\exists$x B(x)

= Vx A(x)   V   $\exists$x B(x)

= RHS

So, true.
Also try using example.
thanks sushant,

if i take one example,we take domain as {3,6,9}

A=div by 2 and B = div by 3

then RHS is obviously TRue but what about LHS --∃x( A(x) -> B(x) ) as here,A is not true for anyone.so LHS is true or false/?
x=6, A is true and B is true. So, LHS=true
sorry,domain is {3,9,15},,then tell for LHS.
Yes, u got it :)
thanks sushant..:-)


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