It is easy to prove that an apple is an apple: The equation x=x is a tautology, for all x. So if x=apple, we can substitute it in the equation x=x and get apple=apple, so an apple is an apple.

But how can one prove (using standard deductive logic) that an apple is not an orange, since x!=y is not a tautology for all x,y?
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It's not a *logical truth* that an apple is not an orange, but it's nevertheless *analytic* with respect to the meaning postulate: if predicate P holds uniformly of object a, then for all predicates Q contrary to P, ¬Q(a). For example, if something is uniformly black, then it's not of any color that is contrary to black (say red). Recall that predicates A, B are *contraries* iff they cannot simultaneously be true of an object:

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> **Definition.** Predicates P and Q are *contraries* =def for all x, either ¬P(x) or ¬Q(x).
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Here's how a proof could proceed. We have a meaning postulate, a convention that states that:

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> **P1.** For all P, x: if P(x), then for all Q: if P and Q are contraries, then ¬Q(x).
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Now we begin with the hypothesis that some object a is an A, and that A and O are contraries:

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> **P2.** A(a)  
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> **P3.** Predicates A and O are contraries
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Then we instantiate (P1) with a for x, obtaining:

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> **P4.** For all P: if P(a), then for all Q: if P and Q are contraries, then ¬Q(a)
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We then instantiate (P4) with A for P and O for Q, obtaining:

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> **P5.** if A(a), then if A and O are contraries, then ¬O(a)
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The rest is two applications of modus ponens starting with (P5) and using (P2) and then (P3):

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> **P6.** if A and O are contraries, then ¬O(a)  
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> **P7.** ¬O(a)
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As has been pointed out already, if you are using logic to talk about apples and oranges you have already provided some model (say, *a* is an apple and *o* is an orange - formally speaking you've set up a function from constants in your logical language to objects or types or fruit or whatever). So I'm reasonably comfortable that the following proof is no less rigorous than what you've got already: let *P* (a one-place predicate) be assigned the meaning 'needs to be peeled to be eaten'\*. Then *Po* is true and *Pa* is false, so then clearly ¬(*x* = *y*) holds.

\*obviously if you don't peel your oranges feel free to substitute some other predicate that fits

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To generalise a little, in case the moral of the above isn't clear, we 'prove' that two objects are distinct by providing some property that holds of one but not the other. It's an open question in metaphysics whether distinct objects can share every property (contenders might be the object of 52 cards and the object of the deck, I guess). Within logic, though, if two elements of your domain are distinct but the same predicates hold of them there isn't a way to prove that they are distinct - you could show that rigorously in a quite straightforward way; proof systems for first-order logic use (roughly) the above method for proving inequality.