Here are sample answers to the 1998 final.  Copies of the 1998 final itself were handed out on the last day of class.  If you didn't get one, there are extra copies outside my office door, Seaman 214.

Also, be sure to do the review sheet handed out on the last day of class.


Symbolizations, Sample Answers

1. Annie dances with Danny only if he is neither bashful nor clumsy.

2. Though Annie is bashful, she is not scared of Danny. 3. Danny, who is bashful, dances with Annie provided that she is not scared of him. 4. If Annie is bashful, then she is scared of every man and woman. 5. Men, who are clumsy, are also bashful. 6. Any woman who dances with Danny is not scared of Annie. 7. Some bashful man is scared of every woman who dances with him. 8. No man will dance with any woman who scares him.  
Indirect Truth Tables

1.  Valid
2.  Invalid.  R: F; S: T; T: F
3.  Valid.
4.  Invalid.  P: T; Q: F; R: T; S: T

Expansions

1.  (Ga -> Fa v Ha) & (Gb -> Fb v Hb)    TRUE

2.  (~Ga & Ha) v (~Gb & Hb) -> Fa & Fb    FALSE

3.  (Ga -> (Fa & Ha) v (Fb & Hb)) & (Gb -> (Fa & Ha) v (Fb & Hb))    TRUE

4.  ((Fa -> ~Ga) & (Fb -> ~Gb)) & Hb    TRUE

Countermodels
Remember: these are just sample answers.  There are lots of other possibilities for countermodels to these sequents.

1.  U: {a,b,c}  F: {a}  G: {a,b}

2.  U: {a,b}  B: { }  C: {a}  F: {a,b}
Note: to get a countermodel for this sequent, the extension of B must be empty.

3.  U: {a,b}  F: {a}  G: {a,b}  H: {b}

Proofs

1. (P -> Q) v (R -> S) |- (P -> S) v (R -> Q)
        1 (1)  (P->Q)v(R->S)        .A
        2 (2)  ~((P->S)v(R->Q))     .A
        2 (3)  ~(P->S)&~(R->Q)      .2 DM
        2 (4)  ~(P->S)              .3 &E
        2 (5)  P&~S                 .4 NEG->
        2 (6)  ~(R->Q)              .3 &E
        2 (7)  R&~Q                 .6 NEG->
        2 (8)  P                    .5 &E
        2 (9)  ~S                   .5 &E
        2 (10) R                    .7 &E
        2 (11) ~Q                   .7 &E
        2 (12) P&~Q                 .8,11 &I
        2 (13) R&~S                 .9,10 &I
        2 (14) ~(P->Q)              .12 NEG->
        2 (15) ~(R->S)              .13 NEG->
        2 (16) ~(P->Q)&~(R->S)      .14,15 &I
        2 (17) ~((P->Q)v(R->S))     .16 DM
        1 (18) (P->S)v(R->Q)        .1,17 RAA (2)

2. @x(Gx v Fx -> (Hx -> Bx)), @y((Hy -> Hy & By) -> Ky) |- @x(Gx -> Kx)
      1 (1)  @x(Gx v Fx -> (Hx -> Bx)) .A
      2 (2)  @y((Hy -> Hy & By) ->Ky)  .A
      1 (3)  Ga v Fa -> (Ha -> Ba)     .1 @E
      2 (4)  (Ha -> Ha & Ba) -> Ka     .2 @E
      5 (5)  ~(Ga -> Ka)               .A [for RAA]
      5 (6)  Ga & ~Ka                  .5 NEG->
      5 (7)  Ga                        .6 &E
      5 (8)  Ga v Fa                   .7 vI
    1,5 (9)  Ha -> Ba                  .3,8 ->E
      5 (10) ~Ka                       .6 &E
    2,5 (11) ~(Ha -> Ha & Ba)          .4,10 MTT
    2,5 (12) Ha & ~(Ha & Ba)           .11 NEG->
    2,5 (13) Ha                        .12 &E
    2,5 (14) ~(Ha & Ba)                .12 &E
  1,2,5 (15) Ba                        .9,13 ->E
    2,5 (16) ~Ha v ~Ba                 .14 DM
  1,2,5 (17) ~Ha                       .15,16 vE
    1,2 (18) Ga -> Ka                  .13,17 RAA (5)
    1,2 (19) @x(Gx -> Kx)              .18 @I

Note: Another obvious way to do this proof would be to assume Ga for -> and then ~Ka for RAA instead of assuming ~(Ga -> Ka) for RAA.

3. @x(~Fx & ~Gx -> ~Hx), $x(Hx & ~Gx) |- $xFx

        1 (1)  @x(~Fx & ~Gx -> ~Hx)  A
        2 (2)  $x(Hx & ~Gx)          A
        3 (3)  ~Fa                   A [for RAA]
        1 (4)  ~Fa & ~Ga -> ~Ha      1 @E
        5 (5)  Ha & ~Ga              A [for $E on 2]
        5 (6)  ~Ga                   5 &E
      3,5 (7)  ~Fa & ~Ga             3,6 &I
    1,3,5 (8)  ~Ha                   4,7 -> E
        5 (9)  Ha                    5 &E
     1,5 (10) Fa                     8,9 RAA (3)
     1,5 (11) $xFx                   10 $I
     1,2 (12) $xFx                   2,11 $E (5)

Another way to do the same sequent.

        1 (1)  @x(~Fx & ~Gx -> ~Hx)   A
        2 (2)  $x(Hx & ~Gx)           A
        3 (3)  ~$xFx                  A [for RAA]
        3 (4)  @x~Fx                  3 QE
        3 (5)  ~Fa                    4 @E
        1 (6)  ~Fa & ~Ga -> ~Ha       1 @E
        7 (7)  Ha & ~Ga               A [for $E on 2]
        7 (8)  ~Ga                    7 &E
      3,7 (9)  ~Fa & ~Ga              5,8 &I
    1,3,7 (10) ~Ha                    6,9 ->E
        7 (11) Ha                     7 &E
      1,7 (12) $xFx                   10,11 RAA (3)
      1,2 (13) $xFx                   2,12 $E (7)

5. @x$y(Fx & ~Gy), @x(Hx -> Gx) |- ~@x(Fx -> Hx)  A looping problem

      1 (1)  @x$y(Fx & ~Gy)     A
      2 (2)  @x(Hx -> Gx)       A
      3 (3)  @x(Fx -> Hx)       A [for RAA]
      1 (4)  $y(Fa & ~Gy)       1 @E
      5 (5)  Fa & ~Gb           A [for $E on 4, b]
      5 (6)  Fa                 5 &E
      3 (7)  Fa -> Ha           3 @E
    3,5 (8)  Ha                 6,7 ->E
      2 (9)  Ha -> Ga           2 @E
  2,3,5 (10) Ga                 8,9 ->E
      1 (11) $y(Fc & ~Gy)       1 @E
     12 (12) Fc & ~Ga           A [for $E on 11, a]
     12 (13) ~Ga                12 &E
 2,5,12 (14) ~@x(Fx -> Hx)      10,13 RAA (3)
 1,2,12 (15) ~@x(Fx -> Hx)      4,14 $E (5)
    1,2 (16) ~@x(Fx -> Hx)      11,15 $E (12)

You are not responsible for proofs like 4, i.e. proofs with two-place predicates.