Running head: PAN-EUROPA
1
Pan-Europa
Amara Fofana
Columbia Southern University
Project Management Strategy and Tactics
MBA 6931
Dr. Jan Tucker
February 27, 2017
PAN-EUROPA
2
References
Meredith, J. R
Running head: PAN-EUROPA
1
Pan-Europa
Amara Fofana
Columbia Southern University
Project Management Strategy and Tactics
MBA 6931
Dr. Jan Tucker
February 27, 2017
PAN-EUROPA
2
References
Meredith, J. R
Running head: PAN-EUROPA
1
Pan-Europa
Amara Fofana
Columbia Southern University
Project Management Strategy and Tactics
MBA 6931
Dr. Jan Tucker
February 27, 2017
PAN-EUROPA
2
References
Meredith, J. R
Running head: PAN-EUROPA
1
Pan-Europa
Amara Fofana
Columbia Southern University
Project Management Strategy and Tactics
MBA 6931
Dr. Jan Tucker
February 27, 2017
PAN-EUROPA
2
References
Meredith, J. R
Running head: PAN-EUROPA
1
Pan-Europa
Amara Fofana
Columbia Southern University
Project Management Strategy and Tactics
MBA 6931
Dr. Jan Tucker
February 27, 2017
PAN-EUROPA
2
References
Meredith, J. R
126
First-order logic
Formal proof
x (x)
x (x)
Let a satisfy (a)
(a)
x (x)
(1)
Given
(2)
Assumption
(3)
(4) -Elimination
(5)
(6) -Elimination
(7)
RAA
Exercise 9.15 Prove that x (x) x (x). (Use -Elim
128
First-order logic
Formal proof
x y (P(x) P(y) (x = y)
x (P(x) R(x)
Let x be arbitrary
R(x)
Let a satisfy P(a) R(a)
P(a)
R(a)
P(x)
(P(x) P(a)
y (P(a) P(y) (a = y)
(P(a) P(x) (a = x)
(a = x)
R(x)
124
First-order logic
!
g(s)sin(x + y)s) ds.
be changed first to something different: f (x + y) =
We will see later how such a change of variable can also be achieved in the
proof system we are study
9.1 First-order languages
123
Formal proof
Let x be arbitrary
(1)
.
(x)
x (x)
(2)
(3) -Introduction
An instance of -Elimination looks like the following.
Formal proof
x (x)
Let a satisfy (a)
(1)
(2)
134
First-order logic
As before, ! is vacuously true if there are no L-structures M making
M ! true. In this case, ! for any whatsoever.
Theorem 9.25 (Soundness) Let L be a first-order language, a set
9.1 First-order languages
133
each function symbol f of arity n in L (so, in particular M is closed under this
function), and there is a relation R M m for each relation symbol R of L of
arity m.
As i
9.1 First-order languages
125
See Exercise 6.22. Alternatively (and Exercise 6.22 asks you to show that this
really is equivalent) you may pretend that there is actually no extra symbol for
implies at
132
First-order logic
As is often the case with formal proofs, there is rarely a single proof for a
particular statement, and in this case a quite different alternative can be given.
Formal proof
a x
9.1 First-order languages
131
Formal proof
a x ( (x) (a)
b y ( (y) (b)
Let a satisfy y ( (y) (a)
Let x be arbitrary
(x) (a)
x ( (x) (a)
a x ( (x) (a)
a x ( (x) (a)
b y ( (y) (b)
(1)
Assumption
(2)
9.1 First-order languages
127
The rest of the proof involves propositional logic and the -Introduction and
-Elimination Rules.
Formal proof (continued)
.
R(a)
R(b)
(a = b)
P(a)
R(a) P(a)
x (R(x) P(x)
130
First-order logic
To see this, suppose t, s are terms and we have proved (t = s). Let (u, v)
be the formula (u = v). Then (t = t) is provable directly using the Reflexivity
Rule. Think of (t = t)
10.1 Proof of completeness and compactness
141
ing of H(L) together with all (x) for formulas and variables x in the first
Henkinisation.
The complete Henkinisation of L is the first-order language LH
9.3 Second- and higher-order logic*
137
Exercise 9.34 Use the previous exercise and induction to show that every
formula (x1 , . . ., xk ) in some first-order language L is equivalent to a formula
in
150
Completeness and compactness
We are going to study the set of all sets of L-sentences of the form Th M
where M is an L-structure. We define
X = cfw_Th M : M is an L-structure .
We will define a to
142
Completeness and compactness
Lemma 10.5 Let L be a first-order language, and suppose is a consistent
set of first-order L-sentences, and is a further L-sentence. Then at least one
of cfw_ , cfw_
148
Completeness and compactness
Then, in contrast with the last example, the set of cyclic groups is not axiomatisable, i.e. there is no set of first-order sentences true for precisely the cyclic
gro
138
First-order logic
Exercise 10.17 for a hint.) In contrast, some cases can be done: we can describe all simple groups of order 168 (or any other finite order) by a first-order
sentence, for example
10.4 The Omitting Types Theorem*
155
V A U. The set A is co-meagre if there are countably many co-rare sets Ai
!
such that A
i=0 Ai .
In the case of our space T where the topology is given by the set
136
First-order logic
(iii) The graph has a clique (complete subgraph) of 5 or more vertices.
(iv) The graph is not complete (i.e. not every pair of vertices is connected
by an edge.)
Exercise 9.32 Wr
10.1 Proof of completeness and compactness
143
s t. Also if t, s, r T and t s and s r then t = s, s = r + so + t = r
hence t r. Thus is an equivalence.
Write [t] for the equivalence class of t T . We
154
Completeness and compactness
In a small number of very special cases in certain languages L and base sets 0
we can build an L-structure M ! + directly from the set + Tmax , often using
an argument
156
Completeness and compactness
The real power of Baires Theorem is that it shows that any countable intersection of co-meagre sets is also co-meagre and hence non-empty. To see this,
!
let Ai be co-
164
Model theory
Proposition 11.8 Let L be a countable first-order language. Then there are
countably many finite strings of symbols from L.
Proof We use the bijection p: N N N of Proposition 11.6, an
10.1 Proof of completeness and compactness
145
This shows that (x) is a t T such that (t) + , so M ! (t) by the
induction hypothesis. Hence M ! x (x), as required.
The case of the quantifier is simila