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I'm in math proof and problem solving and would like someone to tell me if I am on track with these answers. The question is:

Are these statements true or false? The universe of discourse is the set of all people, and T(x, y) means “x and y are twins.” Explain how you determined your solution.

a. ∀x ¬T(x, x) True: For all people there is not a situation where x and x are twins.

b. ∃y T(y, y) False: There exists a y where y and y are twins.

c. ∃x ∀y T(x, y) True: There exists an x that for all y's, x and y are twins.

d. ∀x ¬∃y T(x, y) False: For all x's there does not exist a y where x and y are twins.

e. ∃x ¬∃y T(x, y) True: There exists an x where there is not a y where x and y are twins.

f. ∀x ¬∀y T(x, y) False: For all x's and not for all y's, x and y are twins.

S.E.G.
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  • Can you provide short explanations? For instance, how did you conclude that (c) is true? – bedune4 Jun 28 '16 at 12:41
  • The answers of (c) and (e) are inconsistent, I think too that (c) may be incorrect. But the others look pretty good. – Fernando Jun 28 '16 at 12:53
  • @Fernando The answers of (c) and (e) are not inconsistent. – bedune4 Jun 28 '16 at 13:37
  • @user119394, maybe I'm wrong, but I will tell you my reasoning and then you can help me: suppose there exist an $x$ such that $\forall y(\neq x \mbox{ I think}),;T(x,y)$ (statement (c)), lets call it $x_0$. Then, $\forall x\neq x_0$, (doing $y=x_0$) we have $T(x,y)$, and if $x=x_0$, for any $y$, we have $T(x,y)$. Hence, the validity of (c) implies that $(e)$ is false. – Fernando Jun 28 '16 at 13:56
  • @Fernando Yes you are correct. And that is because $T(x,y) \iff T(y,x)$. I hadn't noticed that (obvious) detail before. Thank you! – bedune4 Jun 28 '16 at 14:23
  • Thank you for the feedback. I read c as "There exists someone (x) who has a twin (y)" which would be true. – S.E.G. Jun 29 '16 at 11:59

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