MATH 10450 TEST 1 STUDY GUIDE
(1) Astronomy
(a) Know the story of how Eratosthenes measured the radius of the earth.
See p.9 of the textbook.
(b) Know the denition of stellar parallax and its applications to measuring
the distance to celestial bodies. p.2
8.4 The Stone Representation Theorem*
115
Given H A let G = cfw_h(a) : a H so H cfw_a A : h(a) G, and if h is
one-to-one then for each a H there is a unique b G such that b = h(a), hence
is onto. If h is onto and G1 , G2 are such that (G1 ) = cfw_a A : h
8.2 Examples and exercises
107
Theorem 8.17 (Boolean Prime Ideal Theorem) Let B be a boolean algebra
and suppose F ! B is a proper filter. Then there is an ultrafilter G F of B.
Proof Let X = B considered as a set of letters, and consider the valuation
v:
88
Valuations
the other hand if max then max by maximality of max and v() =
so v( ) = v() = . This completes the induction step for .
If has n + 1 symbols and is ( ) then by the induction hypothesis
v() = if and only if max and v( ) = if and only if max
7.3 The complexity of satisfiability*
97
!
if ( ) ( ) is not in SAT. The algorithms described earlier can easily be
adapted to solve SAT and TAUT, with similar difficulties about the time that
they take.
In complexity theory, the time taken by an algorith
7
Valuations
7.1 Semantics for propositional logic
Following the general method for other formal systems in this book, we must
connect the system for propositional logic of the last chapter with the boolean
algebras of the chapter preceding it, by using b
9.1 First-order languages
117
We also require symbols for special real numbers such as 0 or 1. We will
want to say when two real numbers are equal, so we will need the symbol =
for equality. We also need to combine real numbers with familiar functions
suc
7.1 Semantics for propositional logic
89
This completes the induction proof and therefore v is a valuation making all
of max (and hence 0 ) true, as required.
Theorem 7.13 (Completeness Theorem, second form) Let X be a set, and
suppose that BT(X) and BT(X
6.1 A system for proof about propositions
69
In this example, brackets were carelessly omitted around the assumption
(a b), and ( a b) in line 3, resulting in an unwanted application of the contradiction rule. (Line 3 is intended to mean ( a) b, though br
8.1 Algebraic theory of boolean algebras
101
xop y = x y
op =
op =
Bop is a boolean algebra in its own right, called the opposite of B. The map
x % x is a homomorphism B Bop which is also one-to-one and onto. In
other words it is an isomorphism of bo
7.2 Examples and exercises
91
Exercise 7.18 Our or or should be interpreted as inclusive or, i.e. a b
is a or b or both. Introduce a new symbol, + for exclusive or, where a + b
means a or b but not both at the same time. Add the axiom
a + b = (a b ) (a b)
104
Filters and ideals
In the sequel, I will tend to concentrate on filters rather than ideals since (being
an optimist) I prefer to focus on true statements rather than false ones, but as
we have seen these ideas are interchangeable. (Is this boolean alg
72
Propositional logic
Formal proof
( )
.
(1) Given
(2) -Elimination
(3) -Elimination
(4)
(5)
Example 6.12 To prove a statement from ( ) and other given statements , deduce both of and first, as shown in the following argument,
and then prove from , and .
8.4 The Stone Representation Theorem*
113
If we now take one step back we may see that this correspondence between the
boolean algebra and its dual space is much deeper and more powerful.
Definition 8.37 Say a topological space X is a Stone space if it is
86
Valuations
of boolean algebras, i.e. any whose valuation is at least as true as that of
however these valuations are chosen, actually has a formal derivation from
assumptions in the formal system. This is hardly an obvious assertion since
the rules fo
114
Filters and ideals
V B, i.e. V X is clopen. Then U = cfw_Dx :V Dx and the inverse image of
U under x # Dx is just V , which is open. Similarly, every open set of a totally
disconnected space is a union of clopen sets. If V X is clopen then the image
94
Valuations
for example p0 = 0, p1 = 1, p2 = 1, p3 = 0, . . ., pk1 = 0 can be specified as
a single boolean term as for example p0 p1 p2 . . . pk1 .)
We also need to consider an additional set of situations that we know to
be impossible, based on the lo
82
Valuations
Now consider:
a
b
a b ( a b) (b ( a b) (a (b ( a b)
So v(a (b ( a b) is for all valuations.
Finally,
a
b
c
b (a b) (a b) c)
Thus (a b) c) has value for some valuations and for others. We
leave the reader to check the remaining examples.
No
9
First-order logic
9.1 First-order languages
Propositional logic is the logic of statements that can be true or false, or take
some value in a boolean algebra. The logic of most mathematical arguments
involves more than just this: it involves mathematica
7.3 The complexity of satisfiability*
99
that SAT P implies that P = NP; in other words, SAT is in some sense the
hardest member of NP (or rather, equal-hardest with some other problems in
NP). Cooks Theorem is proved by describing non-deterministic compu
7.1 Semantics for propositional logic
85
If the last step in p is -Introduction then is ( ) where , have
!
previously been derived from . Thus v(), v( ) ! v() by the induction
!
hypothesis and so v() is a lower bound for both v(), v( ), hence v(
!
) = v
78
Propositional logic
Theorem 6.26 Let X be a set of propositional letters, a finite set of boolean
terms in X, and , , other boolean terms in X. Then:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
if and only if , ;
, if and only if , ;
, if and only if , , ;
, ( ) if
7.1 Semantics for propositional logic
81
on the right and left hand sides of the above equations are interpreted quite
differently, in one case as pure symbols in boolean terms, in the other as
operations in B.
Since BT(X) X, this valuation function BT(X)
68
Propositional logic
We follow with some instructive examples of incorrect proofs.
Example 6.7 Consider the following erroneous proof the shows b is a consequence of (a b). Of course, no such proof should be possible.
Formal proof
(a b)
b
a
b
b
b
(1)
Gi
70
Propositional logic
As with other systems in this book, the proof rules should be used in a
mechanical fashion without any attempt to interpret the symbols with meanings. Thus, for example, a and ( a b) do not together give b directly by
-Elimination.
8.1 Algebraic theory of boolean algebras
105
= []
Proof This is more axiom-checking for energetic students! That is an equivalence is an easy consequence of the proof rules for propositional logic. Note
that the operations are also well defined, for if t
7.2 Examples and exercises
93
Finally, we shall show that with infinitely many propositional letters, our
propositional logic can simulate the systems of Chapters 3 and 4.
Example 7.27 Let X be the set 2 of strings of 0s and 1s of finite length, considere
6.3 Decidability of propositional logic*
77
(Part (c) is possibly too hard at this stage, but will become easy with the results
of the next chapter.)
We have seen many kinds of implication in this book, including the informal
use of implies in our metathe
110
Filters and ideals
and ai Ui = i (U) there is some bi Ui i (A). It follows that the function
b: i # bi is in U A, as required.
We remark that the above proof uses the Axiom of Choice in several different
places. Firstly there is an implicit applicatio
6.1 A system for proof about propositions
71
limited way: to prove a statement it is always possible to start a new subproof
with assumption and try to prove . Our required then follows from
Reductio Ad Absurdum and -Elimination. You will find plenty of o