程序代写代做代考 1

1

Solutions to Predicate Logic Tutorial 3

Q1.

i) c and d.

ii) You have to show ├ cd and ├ dc. I will show the first.

showing ├ cd:

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 cd, discharging 1.

Showing ├ dc :

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 dc, 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.