A.  Symbolize each of the following pairs of Engish sentences and decide whether they are in fact providing exactly the same logical information (disregarding time).  Indicate by "yes" or "no"  and explain how you would know.

1.  Alice will pass if she studies.  Alice will not pass unless she studies.

S -> P  and  ~P v S
No, since they have different truth table values
P
 S
 S -> P
 ~P v S
1
 1
 1
 1
1
 0
 1
 0
0
 1
 0
 1
0
 0
 1
 1
You might also point out that in the first sentence we are saying that study is enough but in the second we are effectively saying it is required.

2.  Alice will pass only if she studies.   Either Alice has studied or she did not pass.

P -> S  and  S v ~P
Yes, since the first expression can be turned into the second through the material implication rule.

3.  If Alice passes she will graduate and she will transfer.   If Alice does not transfer then either she will not have passed or not have graduated.

P  -> (G & T)   and   ~T -> (~P v ~G)
No, since again they have different truth values.
G
 P
T
 P -> (G & T)
 ~T -> (~P v ~G)
1
 1
 1
 1
 1
1
 1
 0
 0
 0
1
 0
 1
 1
 1
1
 0
 0
 1
 1
0
 1
 1
 0
 1
0
 1
 0
 0
 1
0
 0
 1
 1
 1
0
 0
 0
 1
 1
 

4.  Unless Alice graduates and transfers we will know she did not pass.   If Alice did pass then she must have graduated and transferred.

(G & T) v ~P   and   P -> (G & T)
Yes, since using the material implication rule and then the commutation rule we can convert the second expression into the first.

B.  For each of the following pairs of equivalent expressions, present two derivations as in the model above.

1.  A -> (B & C)  and  (~A v  B) & (~A v C)
1.  A -> (B & C)                    \show  (~A v B) & (~A v C)
2.  ~A v (B & C)                    1,  Impl
3.  (~A v B) & (~A v C)           2,  Dist

1.  (~A v B) &  (~A v C)         \show  A -> (B & C)
2.  ~A v (B & C)                    1,  Dist
3.  A -> (B & C)                    2,  Impl                

2.  A <-> B  and  (~A v B) & (A v ~B)
1.  A <-> B                                    \show  (~A v B) & (A v ~B)
2.  (A -> B) & (B -> A)                     1,  Equiv
3.  (~A v B) & (B -> A)                     2,  Impl
4.  (~A v B)  & (~B v A)                    3,  Impl            yes, steps 3 and 4 could be combined
5.  (~A v B) & (A v ~B)                     4,  Com

1.  (~A v B) & (A v ~B)                     \show  A <-> B
2.  (A -> B)  &  (A v ~B)                   1,  Impl
3.  (A -> B)  &  (~A -> ~B)                2,  Impl
4.  (A -> B)  & (B -> A)                     3,  Trans
5.  A <-> B                                      4,  Equiv