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Alex likes anything that contains chocolate.

a - Alex

L(x,y) - x likes y

C(x) - x contains chocolate

$1. \forall x \space (C(x) \implies L(a,x)) $

$2. \forall x \space (C(x) \space \text{^} \space L(a,x)) $

Is there a difference between 1 and 2? I know that the truth tables are different. Is one an incorrect representation in predicate logic?

Edit:

I've removed the truth table as it was completely bogus for predicate logic. I'm still confused with using the $\forall$ vs $\exists$ with an implication. Would $\exists x \space (C(x) \space \text{^} \space L(a,x)) $ be a correct representation of "There exists something that contains chocolate and Alex likes it"? Is this equivalent to "Alex likes anything that contains chocolate"?

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  • $\begingroup$ The second one says in particular $\forall xC(x)$. Can you see why this isn't right? $\endgroup$ – Git Gud Dec 9 '14 at 1:25
  • $\begingroup$ Your truth table isn't correct for doing predicate logic. The statement C(x) (or L(a,x)) is not truth-functional, ie it is not true or false absolutely. To get statements which ARE true or false you first have to define a universe, say the set {a,b} and THEN consider whether statements like C(a) and C(b) are true or false. Just think about the English statement "x contains chocolate." Is that true or false? Depends on what 'x' is! $\endgroup$ – Kevin Driscoll Dec 9 '14 at 1:39
  • $\begingroup$ Ah! I knew it. While I was drawing out the table I had a nagging feeling that this was all wrong. Lets forget about the table. I'm still confused with the 2 expressions. Do all predicate logic statements need an implication? $\endgroup$ – Zee Dec 9 '14 at 1:43
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Proposition #1 is the correct representation of the English sentence. In literal terms it says that "For all objects, if it contains chocolate, then Alex likes it."

Another way of stating #1 symbolically is $\forall x \neg(C(x) \land \neg L(a, x))$ which is equivalent by deMorgan's Laws to $\forall x (\neg C(x) \lor L(a, x))$, which translates to "For any object, either it contains chocolate or Alex likes it."

Proposition #2 does not represent the English statement correctly, saying that "For all objects, it contains chocolate, and Alex likes it." This is clearly false since not all objects contain chocolate (although that would be nice).

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  • $\begingroup$ Hmm.. I thought as much as well. Proposition 2 was given as the answer in a text that I'm studying from. It threw me off for quite a while. So theres no way to represent that statement in any other way (without using an implication)? I'm still a little confused. $\endgroup$ – Zee Dec 9 '14 at 1:36
  • $\begingroup$ There are many ways to represent implication. I've added a few. $\endgroup$ – Vortico Dec 9 '14 at 1:47
  • $\begingroup$ Ah. Thank you, that really cleared things up. :) $\endgroup$ – Zee Dec 9 '14 at 1:50
  • $\begingroup$ Would ∃x (C(x) ^ L(a,x)) be a correct representation of "There exists something that contains chocolate and Alex likes it"? Is this equivalent to "Alex likes anything that contains chocolate"? $\endgroup$ – Zee Dec 9 '14 at 2:02
  • $\begingroup$ No, the existence symbol requires existence of an object satisfying its following condition. Suppose nothing contains chocolate. The statements would then evaluate to two different answers since the original proposition would be vacuously true, while your proposition would be false. $\endgroup$ – Vortico Dec 9 '14 at 2:11

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