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       Andreas van Cranenburgh 0440949
       Core Logic
       Tue Nov 24 13:51:57 CET 2009

Exercise 18
1) i = infinity, because the limits of A and T as they go to infinity 
approach 2.
Achilles is still at 2 at time infinity + infinity, 
because infinity + i = infinity again. 

2)
n = 0. T_0 and T*_0 are both defined to be 1.
A_1 = A_0 + |T_0 - A_0|
A_1 = 0 + |1 - 0| = 1
A*_1 = A_0 + 1/(2^v(0))
A*_1 = 0 + 1/1 = 1

n = 1.
T_1 = T_0 + |T_0 - A_0|
T_1 = 1 + 1/2 * |1 - 0| = 1.5
T*_1 = T_0 + 1/(2^v(1))
T*_1  = 1 + 1/2 = 1.5

A_2 = A_1 + |T_1 - A_1|
A_2 = 1 + |1.5 - 1| = 1.5
A*_2 = A_1 + 1/(2^v(1))
A*_2 = 1 + 1/2 = 1.5

And so on. The formulas can be described by this single formula
(except that for Achilles the sequence is prefixed with 0):

Let i = 1 till infinity: Sum(1/(2^i))

A*_infinity+5 = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375
First we take the result of A*_infinity, which converges to 2. Then we
re-iterate the cycle 5 times to arrive at infinity + 5.

T*_infinity+12 = 2 + 0.5 + 0.25 + 0.125 + 0.0625 + ... + 0.00024 = 2.9997...
Same as before, but now we skip 1/2^0=1 because the turtle's formula increments
the value v(i) by one.

A*_infinity+infinity = 4, Achilles is at 4 at time infinity + infinity, because
once Achilles passes the Tortoise, the sequence starts again and converges to 2
again, and 2 + 2 = 4.
T*_infinity+infinity = 3, similar to Achilles, but since the turtle skips the
step with 1/2^0 = 1, the turtle is behind. This makes sense because the turtle
is slower than Achilles by a factor of 1/2.


Exercise 19

We take L* to be an extension of L, with a new constant .c.
We add the relation greater than to our language:
x > y :=  there is a z: z =/= 0 ^ x = y + z

We define S as the set containing the following formulas:
.c > 0
.c > 1
.c > 1 + 1
.c > (1 + 1) + 1
...

So .c is bigger than all the natural numbers, because it is bigger than all
the successors of zero. Let T' = T U S. Every finite subset of T' can be made
true in some model M, because it contains some sentences of T, and a finite
amount of sentences of the form .c > k.  We can interpret .c as a number larger
than the largest k such that .c > k is in the subset.

Now by the compactness theorem it follows that the whole set T' is true in
some model M. This model M will contain all the natural numbers, because of
T, but it also contains .c, which is not a natural number, thus it is not
isomorphic to N.

Exercise 20

1) Consistent. Model: I(R) = { <1,2>, <1,3>, <2,3> }
2) Consistent. Model: I(R) = {}
3) Inconsistent. If we assume a domain with at least two individuals, then
phi_ii requires there to be a relation between them. This relation satisfies
the antecedent of phi_LEP, which means ~phi_ME is contradicted.
We must assume a domain with at least two individuals because of phi_ME.
Suppose we have a domain with one individual, then R does not hold for
this individual because of phi_i, but this individual will be equal to all
other individuals in the domain. However, ~phi_ME says that neither may hold.

Exercise 21

1) Because with the axiom of separation we have to start from a specific set.
With the axiom of comprehension we were able to implicitly start from the
universal set. Using the axiom of separation it is not possible to arrive
at the universal set, because this axiom is much more restricted, as is shown
by the need for ZFC to add axioms for the existence of the union and powerset
of sets.

2) We substitute the universal set V for X in the axiom of separation:
R = {z in V : phi(z)} where phi(z) = z not in z
To obtain the Russell set R. Now let's ask whether R is in R:
Suppose R is not in R, then by phi it should be in R.
Suppose R is in R, then by phi it shouldn't be in R.
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