1
Solutions to Predicate Logic Tutorial 3
Q1.
i) c and d.
ii) You have to show ├ cd and ├ dc. I will show the first.
showing ├ cd:
1. X (banker(X) estate_agent(X) unpopular(X)) assume
2. banker(a) assume
3. banker(a) estate_agent(a) 2, I
4. banker(a) estate_agent(a) unpopular(a) 1, E
5. unpopular(a) 3,4, E
6. banker(a) unpopular(a) 2,5, I
7. X (banker(X)unpopular(X)) 6,I
In an almost identical way you can show
X (estate_agent(X)unpopular(X))
Then use I to derive
X (banker(X)unpopular(X)) X (estate_agent(X)unpopular(X))
Then by I you get cd, discharging 1.
Showing ├ dc :
1. X (banker(X)unpopular(X)) X (estate_agent(X)unpopular(X)) assume
2. X (banker(X)unpopular(X)) 1, E
3. X (estate_agent(X)unpopular(X)) 1, E
4. banker(a) estate_agent(a) assume
5. banker(a)unpopular(a) 2, E
6. estate_agent(a)unpopular(a) 3, E
7. unpopular(a) Proof by cases, 4, 5, 6
8. banker(a) estate_agent(a) unpopular(a) I, 4, 7
9. X (banker(X) estate_agent(X) unpopular(X)) I, 8
Then by I you get dc, discharging 1.
Q2.
a.
1. X (p(X) q(X) r(X)) given
2. p(a) assume
3. p(a) q(a) r(a) 1, E
4. q(a) r(a) 3, E
5. q(a) 4, E
6. p(a) q(a) 2,5, I
7. X (p(X) q(X)) 6, I
Similarly we prove
X (p(X) r(X))
And then apply I to get:
X (p(X) q(X)) X (p(X) r(X))
b.
1. X (p(X) (q(X) r(X))) given
2. p(a) q(a) assume
3. p(a) 2, E
4. q(a) r(a) 1,3, E
5. q(a) 2, E
6. r(a) 4, 5,E
7. p(a) q(a) r(a) 2,6, I
8. X (p(X) q(X) r(X)) 7, I
c.
1. X (p(X) ¬q(X)) given
2. p(a) given
3. Y(q(Y) s(Y)) given
4. ¬q(a) 1,2,E
5. q(a) s(a) 3, E
6. s(a) 4,5, E
d. Hint: Think of using proof by cases. Then it is easy.