I've thought about it into two ways but I don't know if either, neither or bow are correct:

First using sets: Given the sets "T" (tall people) y "R"(rich people), for Joe to be tall and rich, Joe has to belong to the intersection of T and R, so, he has to belong to both T and R. Let's suppose he does.

I understand that the proposition is saying that for all "t" belonging to "T", "t" doesn't belong to R (no tall person is rich). Then if Joe belongs to T (is tall), he can't belong to R(be rich), and then if he can't belong to R ,he can't belong to R&T intersection either, which contradicts with the initial supposition, so he can't be tall and rich simultaneously.

Second approach:

t: Joe is tall
r: Joe is rich

Is the function well formed? if not, why?

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So, 1) is any of my approaches valid? 2) If not, how can I prove or disprove this?:

None of the tall people are rich. Joe is rich, therefore Joe isn't tall."

  • It looks like you're using both A and T for the set of tall people, but you can definitely do this with set theory. Your second approach may also work, but I have the impression that set theory is a closer formalisation of the text.
    – user2953
    Jun 10 '15 at 6:34
  • @Keelan you're right, I translated it from Spanish and I made some mistakes, thanks for noticing, I will edit it. Jun 10 '15 at 6:37
  • I wouldn't say the translation is "invalid" (meaning "wrong") per se, but I'd say the translation is kind of stilted. Sentential logic is not well suited for "all" and "none" claims compared to set approaches.
    – virmaior
    Jun 10 '15 at 6:44
  • In the truth table you don't need r-> ~t and its contraposition, one is enough. Jun 10 '15 at 9:27
  • Your "set argument" works ... Jun 10 '15 at 9:51

The coditional is "inside" a quantifier :

there is no x (Tall(x) and Rich(x)),

that is equivalent to :

forall x not (Tall (x) and Rich(x)).

Thus, if we "instantiate" the universal quantifier with John, we get :

not (Tall(John) and Rich(John)).

Bur Rich(John) holds, and thus we have to conclude with : not Tall(John).


Your first approach corresponds roughly to Venn's diagrams, and it's valid.

Your second approach could be made more rigorous. You'd have to start from predicate logic (not propositional logic):

(~Exists x) Tx & Rx

Or equivalently:

(Forall x) Tx -> ~Rx

Then apply this to John and you get:

Tj -> ~Rj

Now if you call t the proposition Tj and r the proposition Rj you fall back on your solution (except that you don't need its contraposition) and it's valid.

Also, you can do without truth tables to have a more rigoreous demonstration.

Tj -> ~Rj

Implies, by contraposition:

Rj -> ~Tj

Add premiss:


And with modus ponens you get:


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