Prenex expressions

are WFFs in which all the quantifiers are lined up on the left.  Another way of saying this is that the first quantifier is the main connective.  Converting forms and showing their equivalence are excellent exercises for practice.

Examples

(Ex)Fx v (Ex)Gx |- (Ex)(Fx v Gx)

1.  (Ex)Fx v (Ex)Gx      \(Ex)(Fx v Gx)
      |  2.    ~(Ex)(Fx v Gx)     AIP    we are going to use an indirect derivation
      |  3.     (x)~(Fx v Gx)      2, CQ
      |  4.      (x)(~Fx & ~Gx)  3,  DM
      |  5.       ~Fa & ~Ga        4,  UI
      |  6.       ~Fa                     5,   Simp
      |  7.       (x)~Fx                6,  UG   note what we need to do in order to move ahead
      |  8.        ~(Ex)Fx            7,  CQ
      |  9.       (Ex)Gx              1,8,  DS
      | 10.      ~Ga                   5,  Simp
      | 11.       (x)~Gx              10,  UG
      | 12.       ~(Ex)Gx            11,  CQ   again we need to get the opposite of Ex(Gx)  
      | 13.       (Ex)Gx & ~(Ex)Gx      9,12  Conj
14.   (Ex)(Fx v Gx)             2-13  IP

(Ex)Fx -> (Ey)Gy |- (x)(Ey)(Fx -> Gy)

1.  (Ex)Fx -> (Ey)Gy    / (x)(Ey)(Fx -> Gy)
     |  2.   Fa                     ACP   we are going to use a conditional derivation
     |  3.   (Ex)Fx              2,  EG
     |  4.   (Ey)Gy              1,3,  MP
     |  5.    Gb                    4,  EI         (y=b)
6.    Fa -> Gb           2-5,  CP
7.  (Ey)(Fa -> Gy)           6,   EG
8.  (x)(Ey)(Fx -> Gy)      7,   UG   remember this is a legitimate step even though Fa was assumed; the truth value of the conditional is not affected even if there were no one who had the characteristic "F"

(x)(Ey)(Fx -> Gy) |- (Ex)Fx -> (Ey)Gy

1.  (x)(Ey)(Fx -> Gy)    (Ex)Fx -> (Ey)Gy
     |  2.   (Ex)Fx             ACP
     |  3.   Fa                     2,  EI      (x=a)
     |  4.   (Ey)(Fa -> Gy)    1,  UI
     |  5.    Fa -> Gb            4, EI    (y=b)
     |  6.    Gb                      3,5,  MP
     |  7.    (Ey)Gy               6,  EG
8.  (Ex)Fx -> (Ey)Gy     2-7  CP


Here are the most common equivalences.  A good practice is to do the derivation for each.

(x)Fx & (x)Gx is equivalent to  (x)(Fx & Gx)
(Ex)Fx v (Ex)Gx  is equivalent to  (Ex)(Fx v Gx)
(x)Fx -> P  is equivalent to   (Ex)(Fx -> P)
(Ex)Fx -> P  is equivalent to   (x)(Fx -> P)
P -> (x)Fx  is equivalent to   (x)(P -> Fx)
P -> (Ex)Fx  is equivalent to   (Ex)(P -> Fx)