Quiz 5 ANSWERS
Note: there are several ways to do each of the derivations.
Derive the conclusions:
1. A  (~ B V C)
/ A C
2. B
3. A
ACP
4. ~B V C
1, 3 MP
5. ~ ~B
2, DN
6. C
4, 5 DS
7. A  C
3 – 6, CP
(Note: To use CP efficiently here requires IP as well. First assume A
for CP and then ~C for IP; it is fairly easy after that.)
1. (J • S) V B
2. (~ B  S)  E
/ E
3. B V (J • S)
1, Com
4. (B V J) • (B V S)
3, Dist (Note: the critical step)
5. B V S
4, Simp
6. ~~B V S
5, DN
7. ~ B  S
6, Imp
8. E
2, 7 MP
(Note: this goes well as an IP, although it is longer. Assume ~E, you
will eventually get S •~S.)
1. ~ B
2. C  (~ D• B)
/~ C
3. C
AIP
4. ~ D• B
2, 3, MP
5. B
4, Simp
6. ~ B• B
5, 1, Conj
7. ~C
3-6, IP
(Note: One way to do this as a direct proof is to make step #3 the
addition ~~D V ~B; then do a DeMorgans to get ~(~ D• B) and ~C
follows in one step. Or, use Imp on line 2 and then Dist, which, after
a few steps, gives you C  B. Then MT with #1, gives ~C. All
equally correct.)
1. A  ~ B
/ ~A 
2. (A • ~B) V (~A • ~~B)
B
1, Equiv
3. (~A • ~~B) V (A • ~B) 2, Com
4. (~A • B) V (A • ~B)
3, DN
5. (~A • B) V (~~A • ~B)
4, DN
6. ~A 
5, Equiv
B
(Note: this can also be done using the other Equiv rule, although it
will be longer. Or it can be done as two separate CPs: first to get ~A
 B and then to get B  ~A. This is longer as well.)