[Discrete Maths] Predicate Logic):-

Question

A(x) :- Apple on the table.

Give predicate logic that there is at most one apple on the table.

1. ∃x∃y(A(x) ^ A(y) ) ->x=y

2.∀x∀y(A(x) ^ A(y) ) ->x=y

I know first one is correct,but why second one is not correct?

If i translate first:- If there exists two apples then they must be same.And if LHS is false,i.e there is no apple then RHS will be true.So it fits both cases of 0 and 1 apple

If i translate second:- If any two objects in the universe are apples on the table ,then they must be same.It also follows same thing.

So are both versions correct?Or i am mistaking somehwere?

Answer 1

 predicate logic that there is at most one apple on the table

More than one apple can't be on the table as per the predicate logic

Comments

Please elaborate ?

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In the predicate logic it was clearly said there can be Atmost​ one APPLE on the table independent of how many apples are present in the UOD

Answer 2

I'm going to slightly modify A(x) for better understanding. 

A(x) = x is an apple on the table.

Statement one is correct.

Statement two does not assert - If any two objects in the universe are apples on the table ,then they must be same.

Statement two asserts that - If, every object x in the universe is an apple on the table and every object y in the universe is an apple on the table then x = y.

As a rule of thumb, remember this - There exists goes with conjunction and For Every goes with implication.

Comments

Got it. thanks.I got confused in precedence.

.∀x∀y(A(x) ^ A(y) ) ->x=y ,here implication for each pair of x and y or will it work for all x and y combined ?

I mean will it be :- .( ∀x∀y(A(x) ^ A(y) )  )->x=y

or

.∀x∀y( (A(x) ^ A(y) ) ->x=y)

Please see the brackets in above two?

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.∀x∀y( ( A(x) ^ A(y) ) ->x=y )

Can anyone confirm if the above is correct?