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

As before, I use the @ for the universal quantifier and the $ for the existential quantifier.

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 exactly two bosses. [Corrected 4/17]

$xy((Bx & By) & (~x=y & @z(Bz -> z=x v z=y))

it's also ok to use the non-identity sign

14. There are at least two workers who are not lazy.

$xy(((Wx & ~Lx) & (Wy & ~Ly)) & ~x=y)

it's also ok to use the non-identity sign

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 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))