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Question Title may seems similar but Its not a Duplicate Question.


Question:-

Suppose I want to formulate a statement "All Apples are Delicious.

Let F be the domain of fruits and

A(x) : is an apple

D(x) : is delicious

Let's say:

∀x∈F, A(x)⟹D(x)

Is Correct and means all apples are delicious.

And

∃x∈F,A(x)⟹D(x)

It means there is some fruit that if it is an apple, it is delicious.

It is Incorrect because if x is not an apple then also it is true i.e it also includes other fruits


My Problem:-

Why we can't apply the same logic (that is if x is not an apple then also it is true.) for Universal Quantification (∀x∈F, A(x)⟹D(x))?

Why Only for Existential Quantifier we don't want it to hold true in a universe that doesn't contain any apples why not for Universal Quantifier?


P.S. :- Sorry in advance because my English is not upto that mark. Edits are welcome :)

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  • 2
    $\begingroup$ What do "correct" and "incorrect" mean? "All apples are delicious" is either true or false. The statement is correctly formulated as a statement in the predicate calculus. The second statement is rather odd, but I don't see what justifies you in saying it is incorrect. Perhaps it doesn't express what you want to say, but then you haven't told us what you want to say, have you? $\endgroup$ – saulspatz Feb 17 '18 at 6:24
  • $\begingroup$ @saulspatz I edited the question. I think now the meaning of correct and incorrect are clear. :) $\endgroup$ – Bhaskar Feb 17 '18 at 6:34
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I'm going to answer the question I think you're trying to ask, so correct me if I'm wrong.

First, you want to express the sentence "All apples are delicious," and you correctly say $$\forall x \in F,A(x) \Rightarrow D(x)$$ Note that if all apples are delicious, this statement is true of every fruit. If the fruit is an apple, it's true because the apple is delicious. If the fruit is not an apple, it's true because the hypothesis is false, and the statement is true.

Now I guess you wanted to express the statement there is some apple that is not delicious, which you should have written as $$\exists x \in F,A(x) \wedge D(x),$$ or something equivalent. Instead you wrote $$\exists x \in F,A(x) \Rightarrow D(x)$$ Now, this statement can fail in two ways. First it may be that no apple is delicious. Second, there may be some fruit that isn't an apple. Once again, if the hypothesis is false, the statement is true. That was very convenient in the first case. I want to make a statement that formally applies to every fruit, although I'm really just talking about apples. The convention that an implication with a false hypothesis is always true is just what we need. However, trying to use it in the second case backfires. Remember, to prove an existential statement, it's enough to produce one example. We don't want that to be an example of something were not even really talking about.

I think beginners often find that technicalities obscure what is really quite straightforward. Universal quantifiers at the beginning of a sentence are actually sort of redundant. All you really want to say is "All apples are delicious." With only the predicates $A$ and $D$ at your disposal, what can you say but "if it's an apple, then it's delicious?" The other case, is similar, all you can is, "There is some thing that is both an apple and delicious."

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  • $\begingroup$ For ∀x∈F,A(x)⇒D(x)... If the fruit is not an apple, it's true because the hypothesis is false, and the statement is true. I may happen that there is no apple. Then also this statement i.e. ∀x∈F,A(x)⇒D(x) is true.... While for ∃x∈F,A(x)⇒D(x)... If the fruit is not an apple, it's true because the hypothesis is false, and the statement is true. It may happen that there is no apple. Then also this statement i.e. ∃x∈F,A(x)⇒D(x) is true.... then why we should not ∃x∈F,A(x)⇒D(x) while we can use ∀x∈F,A(x)⇒D(x)? $\endgroup$ – Bhaskar Feb 17 '18 at 7:31
  • $\begingroup$ Yes, if all the fruits are apples, then the the statement is correct, but then you don't even need the hypothesis. You could say $\exists x \in F, D(x).$ Still, neither of these is the right answer. Your job is not to find an expression that might be true under some circumstances; you are to find a formal statement that correctly models the everyday statement in all circumstances. Once again, to prove you have not done that, all I have to do is show you one counterexample. $\endgroup$ – saulspatz Feb 17 '18 at 7:37
  • $\begingroup$ ∀x∈F, A(x)⟹D(x) that is for all fruits if fruit is Apple then its Delicious. Now by Implication Logic we all know this If hypothesis is false then also Implication is True by this we can say For all fruits if fruit is not an apple then its Delicious/not Delicious both are True I want to ask how "For all fruits if fruit is not an apple then its Delicious/not Delicious" and "All Apples are Delicious" are similar or say convey the same meaning? $\endgroup$ – Bhaskar Feb 17 '18 at 9:14
  • $\begingroup$ If you have a new question, ask a new question $\endgroup$ – saulspatz Feb 17 '18 at 15:21
  • $\begingroup$ Its similar to my previous question, So I thought If I ask then it may be mark as duplicate $\endgroup$ – Bhaskar Feb 17 '18 at 15:39

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