T9.1 3 of 4

Let's continue to use our symbolization key from the last page. Our universe of discourse is the class including Sam (s) and Chris (c). We will want to symbolize sentences referring to the seniors in class (Sx: x is a senior) and to the females (Fx: is a female). In addition we continue to use 'I' for identity ('Ixy' means "x=y").

Now, how should we symbolize

(6) There is exactly one senior in class.

Part of what (6) says is that there is a senior in class: '(%x)Sx'. But (6) also says that this person is the only senior, that no one other than x is a senior:
    ~(%y)( y is distinct from x & y is a senior).
Or, to put the two parts together in hybrid form:

(%x)[ x is a senior & ~(%y)( y is distinct from x & y is a senior ) ]

Which leads to a

Symbolization for (6):                     
(%x)[Sx&~(%y)(~Iyx&Sy)]

Another way to think about about (6) is

(%x)[ x is a senior & everyone who is a senior is equal to x ]

Notice that the second conjunct within the scope of the existential quantifier again says that no one other than x is a senior. So,

A second symbolization for (6):               
(%x)[Sx&(^y)(Sy>Iyx)]

And a third way to do this is simpler:

A third symbolization for (6):               
(%x)(^y)(Sy=Iyx)]

Makes sense? Similarly, for

(7) Exactly one female is a senior.

we write:

(%x)[ x is a female senior & no one else is a female, senior ]

or

(%x)[ (Fx&Sx) & ~(%y)( y is distinct from x & y is a female, senior ]

which in pure PL is:

Symbolization for (7):
(%x)[ (Fx&Sx) & ~(%y)( ~Iyx & (Fy&Sy) ]

or we could do it the simpler way with the triple-bar:

A second symbolization for (7):
(%x)(^y)[ (Fy&Sy)=Iyx ]


One last example:

(8) At most one female is a senior.

This means that there are not two. As far as (8) is concerned, there could be no females who are seniors or there could be one. But no more. So, we could symbolize (8) into hybrid form as:

(There are no female seniors) v (there is exactly one)

i.e., using our ideas from above:

A symbolization for (8):                       
~(%x)(Fx&Sx) v (%x)[(Fx&Sx)&(^y)((Fy&Sy)>Ixy)]

But there is a better way, (8) says that there are no two distinct females who are seniors:

A shorter symbolization for (8):                 
~(%x)(%y)[~Ixy & ((Sx&Sy) & (Fx&Fy))]



  • To finish this page and move on to the tutorial's final page, do the following symbolizations.
  • The "hint" link will fill in the hybrid form for you, in case the answer's not obvious.
  • To finish a problem, convert from the hybrid form to PL.
  1. There is at least one female senior. (Hint)
  2. There are at least two seniors. (Hint)
  3. There is exactly one senior. (Hint)
  4. At most one female is a senior. (Hint)
  5. There is at least one senior other than Chris. (Hint)