Exercise 6.2 - Answers

 

1

(i)

$x[[MxÙ"y[My®x=y]]ÙLxb] (There is exactly one man and he likes Mary.)

 
(ii)

First step: There is exactly one person who likes Mary and he likes everyone.

So: $x[[[HxÙLxb]Ù"y[[HyÙLyb]®x=y]]Ù"z[Hz®Lxz]]

 
(iii)

First step: $x$y[[x is the man who likes Mary Ù y is the woman who likes John] Ù Lxy].

x is the man who likes Mary: [[MxÙLxb]Ù"z[[MzÙLzb]®z=x]]

y is the woman who likes John: [[WyÙLya]Ù"u[[WuÙLua]®u=y]]

So: $x$y[[[[MxÙLxb]Ù"z[[MzÙLzb]®z=x]]Ù[[WyÙLya]Ù"u[[WuÙLua]®u=y]]]ÙLxy]

 

 
(iv)

[[MaÙ"x[Hx®Lax]]Ù"y[[MyÙ"x[Hx®Lyx]]®y=a]]. (Notice that it is all right to use "x twice, because the second occurrence is not in the scope of the first.)

 
(v)

¬$z [HzÙ[[MzÙ"x[Hx®Lzx]]Ù"y[[MyÙ"x[Hx®Lyx]]®y=z]]]. (We don't, of course, want to say that there is something which is the man who likes everyone, and there is no such person as it.)

 

(vi)

This is perhaps ambiguous. It could be committed to the existence of the man who like everyone, and be saying that if John is he, John likes Mary. But (more likely?) it is not committed to the existence of such a man, but is saying that if John is such a man, he likes Mary. Taken the second way we have:

[[[MaÙ"x[Hx®Lax]]Ù"y[[MyÙ"x[Hx®Lyx]]®y=a]]®Lab]

Taken the first way we would have:

$z[[[MzÙ"x[Hx®Lzx]]Ù"y[[MyÙ"x[Hx®Lyx]]®y=z]]Ù[a=z®Lab]]

(Consider instead, "If John is the person who took my milk, he is no friend of mine". Here, most likely, I do mean to commit myself to the existence of the culprit.)

 

(vii)

$xx=x. (Or we could have, say, $x[MxÚ¬Mx]. But using "=" seems neater.)

  (viii)

$xx=x

  (ix)

$x$y¬x=y

 

(x)

$x$y[¬x=yÙ"z[z=xÚz=y]]

2

(i)

"x[Ux®$yTyx]

  (ii)

¬$x[$yTxyÙ$zTzx]. Or, "x[$yTxy®¬$zTzx]. (If you really want to have all the quantifiers at the beginning, you would have ¬$x$y$z[TxyÙTzx]; or, "x"y"z[Txy®¬Tzx]. But there is absolutely no need, and it is easy to make mistakes trying to achieve this bit of regimentation.)

  (iii)

$x[UxÙ$y$z[[TyxÙTzx]Ù¬y=z]]

  (iv)

First step: ¬$x[UxÙ x is a member of more than one college]

x is a member of more than one college: $y$z[[[CyÙMxy]Ù[CzÙMxz]]Ù¬y=z]

So:  ¬$x[UxÙ$y$z[[[CyÙMxy]Ù[CzÙMxz]]Ù¬y=z]]

  (v)

First step: $x$y[[¬x=yÙx and y are members of different colleges]Ù$z[TzxÙTzy]]. (We need "¬x=y", because otherwise what is said could be made true by one person who is a member of more than one college – supposing such a thing to be possible.)

x and y are members of different colleges: $u$v[[[CuÙCv]Ù¬u=v]Ù[MxuÙMyv]]

 

So: $x$y[[¬x=yÙ$u$v[[[CuÙCv]Ù¬u=v]Ù[MxuÙMyv]]]Ù$z[TzxÙTzy]]


  (vi)

First step: ¬$x$y[[¬x=yÙ$u$v[[[CuÙCv]Ù¬u=v]Ù[MxuÙMyv]]]Ù x and y have exactly the same tutors] (The first part is taken from the answer to (v) above.)

x and y have exactly the same tutors: "z[Tzx«Tzy]

 

So: ¬$x$y[[¬x=yÙ$u$v[[[CuÙCv]Ù¬u=v]Ù[MxuÙMyv]]]Ù"z[Tzx«Tzy]]

3

(i)

There is an undergraduate reading Philosophy and Economics and one reading Philosophy but not Economics.

  (ii)

This says that there is exactly one thing which… which what? That is given by [CxÙ¬$y[[MyxÙUy]Ù¬Py]]. This means: x is a college and there is nothing which is a member of it and is an undergraduate and is not reading Philosophy. That is, x is a college, all of whose undergraduate members are reading Philosophy.

So: There is just one college all of whose undergraduate members are reading Philosophy.

  (iii)

This says that, for any x, if there is someone (strictly something) reading Philosophy whose tutor is x, then no one (thing) whose tutor is x is reading Economics. So: No one (thing, strictly) tutors both Philosophy and Economics pupils.

  (iv)

Some tutors are members of more than one college.

  (v)

Sometimes tutors who are members of one college teach pupils who are members of another.

  (vi)

Anyone (thing) who tutors some pupils who are reading Philosophy also tutors some who are reading Economics.

  (vii)

This says that there is exactly one thing of which something is true and that something else is true of it. The first something is given by "z[Pz®Txz], which means: x is tutor to everyone (everything) reading Philosophy. The something else is given by "z[Ez®¬Txz], which means: x is tutor to no one reading Economics. So the whole formula means: there is exactly one thing which is tutor of everyone reading Philosophy and it is tutor of no one reading Economics.

So: The person who tutors everyone reading Philosophy tutors no one reading Economics.

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