Help Wanted Symbolizations

Sample Answers

a: Homer / Bx: x is a boss / Fxy: x fires y
b: Mr. Burns / Wx: x is a worker / Hxy: x hires y
Lx: x is lazy

There will in many cases be other acceptable answers.

1. Homer is a lazy worker.

La & Wa

2. Mr. Burns is not a lazy boss.

~(Lb & Bb)

3. Mr. Burns hires Homer.

Hba

4. Mr. Burns fires some worker.

x(Wx & Fbx)

5. All workers get fired by Mr. Burns.

x(Wx  Fbx)

6. Mr. Burns is a boss who hires some lazy workers.

Bb & x((Wx & Lx) & Hbx)

7. Some bosses fire all lazy workers.

x(Bx & y(Ly & Wy  Fxy))

8. All bosses fire all lazy workers.

x(Bx y(Wy & Ly  Fxy))

9. Some lazy bosses hire only lazy workers.

x((Bx & Lx) & y(Hxy  Ly & Wy))

10. Some lazy bosses hire no lazy workers.

x((Lx & Bx) & ~y((Ly & Wy) & Hxy))
x((Lx & Bx) & y(Ly & Wy  ~Hxy))

11. No worker fires any boss.

x(Wx  ~y(By & Fxy))
x(Wx y(By  ~Fxy))
~x(Wx & y(By & Fxy))

12. There is some worker who Mr. Burns has both hired and fired.

x(Wx & (Hbx & Fbx))

13. There are at least two bosses.

xy((Bx & By) & ~x=y)

xy((Bx & By) & x≠y)

14. There are some workers who are not lazy.

x(Wx & ~Lx)

15. Mr. Burns fires any lazy worker.

x(Wx & Lx  Fbx)

16. Any lazy worker who Mr. Burns hires he also fires.

x((Wx & Lx) & Hbx  Fbx)

17. Neither all workers nor all bosses are lazy.

~x(Wx  Lx) & ~x(Bx  Lx)

~(x(Wx  Lx) v x(Bx  Lx))
x(Wx & ~Lx) & x(Bx & ~Lx)

Note that the following is NOT correct: ~x(Bx v Wx  Lx). This symbolization says something weaker than the original sentence since it's compatible with all bosses being lazy (if at least some workers are not lazy) and likewise is compatible with all workers being lazy (if at least some bosses are not lazy). But the original sentence is not compatible with either of these. For the original sentence to be true there have got to be at least one worker and at least one boss who are not lazy. The proposed symbolization is true when there is at least one worker or at least one worker who is not lazy.

18. Not every worker is lazy.

~x(Wx  Lx)

19. No boss fires himself.

x(Bx  ~Fxx)
~x(Bx & Fxx)

20. A worker’s being lazy is sufficient for him to be fired by some boss.

x(Wx & Lx y(By & Fyx))

21. Mr. Burns hires at least two workers.

xy(((Hbx & Hby) & (Wx & Wy)) & ~x=y)

xy(((Hbx & Hby) & (Wx & Wy)) & x≠y)

22. Mr. Burns hires at least three workers.

xyz(((Hbx & Wx) & (Hby & Wy)) & (((Hbz & Wz) & ~x=y) & (~x=z & ~y=z)))

xyz(((Hbx & Wx) & (Hby & Wy)) & (((Hbz & Wz) & x≠y) & (x≠z & y≠z)))

23. Mr. Burns hires exactly two people.

xy(((Hbx & Hby) & (Px & Py)) & z(Pz & Hbz  z=x v z=y))

24. No boss hires anyone who is lazy.

x(Bx y(Hxy & Py  ~Ly))

x(Bx  ~y((Hxy & Py) & Ly))

25. Homer is not the only lazy person.

(La & Pa) & x((Px & Lx) & ~x=a)

(La & Pa) & x((Px & Lx) & x≠a)

26. Homer is not the only lazy worker hired by Mr. Burns.

((La & Wa) Hba) & x((Lx & Wx) & (Hbx & ~x=a))

27. In addition to Homer, who is a lazy worker, there are two more.

(La & Wa) & xy(((Lx & Wx) & ~x=a) & ((Ly & Wy) & (~y=a & ~x=y)))

28. There is at most one boss.

~xBx v (xBx & y(By  x=y))

29. There is exactly one boss in addition to Mr. Burns.

Bb & x((Bx & ~x=b) & y(By  y=x v y=b))

30. There is at most one boss in addition to Mr. Burns.

Bb & (y(By y=b) v x((Bx & ~x=b) & y(By  y=x v y=b)))