This is a problem from Discrete Mathematics and its Applications enter image description here

My question is on 10g. Here is my work so far. enter image description here

My logic behind this is to first iterate over all peoples in the world, for each person in the world, again iterate over all peoples in the world. This is going to find a combination of two unique people(bc of x != y) that Nancy can fool. How can i put a quantifier or check to make sure that this is the only combination? After all, one the main tools that I was given enter image description here represents "there is". It doesn't check exact amount(in this case 2). I don't what this would be called but I tried looking up exact quantifiers(a logical guess) - https://www.google.com/#q=exact+quantifiers but didn't find anything on this subject - it was all stuff on regular expressions, php, regex, asymmetrics, etc.


The way you add the fact that x and y are unique is you add a universal quantifier that if anything else satisfies the relationship, then it must be x or y. So you'd add an $\wedge (\forall w, F(Nancy,w) \implies ((w=x)\vee (w=y))$ to the end of what you have

  • $\begingroup$ would that be another nested quantifier? $\endgroup$ – committedandroider Jan 10 '15 at 23:26
  • $\begingroup$ It would go right after your third and term, before the closing parenthesis $\endgroup$ – Alan Jan 10 '15 at 23:31
  • $\begingroup$ oh because it needs access to x and y $\endgroup$ – committedandroider Jan 11 '15 at 2:25
  • $\begingroup$ Correct, this is the general way you express something is unique...you say that if anything else satisfies the criterea, then it must be equal to the original thing $\endgroup$ – Alan Jan 11 '15 at 2:40
  • $\begingroup$ Hi Alan, can you take a look at my latest question on proof by contradiction? $\endgroup$ – committedandroider Jan 11 '15 at 20:39

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