"Jan Brosius" <[EMAIL PROTECTED]> writes:
> > SORRY, this is quite TRUE , in fact [forall x. alpha(x)] <=> alpha(x)
> >
> > the above true equivalence seems to be easily considered as wrong . Why?
> > Because alpha(x) is TRUE can be read as alpha(x) is TRUE for ANY x.
> >
> > (Is there something wrong with the education of a computer scientist?)
Jan, could you tell us whether you think the following statements are
true or false?
Let prime(x) mean "x is a prime number".
1. [forall x. prime(x)] <=> prime(x)
2. forall x.([forall x. prime(x)] <=> prime(x))
3. forall y.([forall x. prime(x)] <=> prime(y))
4. [forall x. prime(x)] <=> prime(y)
5. [forall x. prime(x)] <=> prime(2)
6. [forall x. prime(x)] => prime(2)
7. prime(2) => [forall x. prime(x)]
8. prime(2)
9. [forall x. prime(x)] <=> prime(4)
10. [forall x. prime(x)] => prime(4)
11. prime(4) => [forall x. prime(x)]
12. prime(4)
13. forall x. prime(x)
14. prime(x)
15. Statement 1 above means the same thing as statement 2 above.
16. Statement 2 above means the same thing as statement 3 above.
17. Statement 3 above means the same thing as statement 4 above.
18. Statement 5 above is a substitution instance of statement 3;
thus, if statement 3 were true, statement 5 would be true.
19. Statement 13 above means the same thing as statement 14 above.
If we follow the convention that free variables are to be considered
implicitly universally quantified, my vote is that statements 6, 8, 9,
10, 11, 15, 16, 17, 18, and 19 are true; the rest are false.
Carl Witty
