Branden Fitelson
Philosophy 12A Notes
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Overview of Today’s Lecture
•
Today’s Music:
Led Zeppelin
•
The midterm is on Thursday, 6/10 (in class)
.
–
I’ve posted (and discussed) a sample midterm. It has the same
structure and complexity as the actual midterm (good study guide).
•
I have posted HW #3, which is due on Thursday @ 4pm in the drop box.
–
It’s all chapter 3 problems — truthtable methods for validitytesting.
•
I posted revised versions of lecture #6 and my “short method” handout.
•
MacLogic
— a useful computer program for natural deduction.
–
You might want to download
MacLogic
at this point .
. .
–
We’ll be using it very soon (and for the rest of the term) .
. .
–
See
http://fitelson.org/maclogic.htm
•
Today: Chapter 3, Finalé and Chapter 4 Intro.
UCB Philosophy
Chapter
3
4
(Intro)
06/08/10
Branden Fitelson
Philosophy 12A Notes
2
'
$
%
Expressive Completeness: Rewind, and More ExtraCredit
•
Q
. How can we deﬁne
↔
in terms of

?
A
. If you naïvely apply the
schemes I described last time, then you get a
187 symbol monster
:
[
p
↔
q
± ,
A

A
, where
A
is given by the following
93 symbol
expression:
(((p

(q

q))

(p

(q

q)))

((p

(q

q))

(p

(q

q))))

(((q

(p

p))

(q

(p

p)))

((q

(p

p))

(q

(p

p))))
•
There are
simpler
deﬁnitions of
↔
using

.
E.g.
, this
43 symbol
answer:
[
p
↔
q
± ,
((p

(q

q))

(q

(p

p)))

((p

(q

q))

(q

(p

p)))
•
I oﬀered E.C. for the shortest solution. Some students have come up
with it (the shortest solution is
<
25 symbols, counting parens).
•
More E.C.
Find the
shortest possible
deﬁnitions of (1)
[
p
→
q
±
, (2)
[
p
∨
q
±
, and (3)
[
∼
p
∼
q
±
in terms of
p
,
q
, and the NAND operator

.
•
If you submit EC, please
prove
the correctness of your solution, using a
truthtable method. You may submit these E.C. solutions to your GSI.
UCB Philosophy
Chapter
3
4
(Intro)
06/08/10
Branden Fitelson
Philosophy 12A Notes
3
'
$
%
Presenting
Your “ShortCut” TruthTable Tests
•
In any application of the “short” method, there are two possibilities:
1. You ﬁnd an interpretation (
i.e.
, a row of the truthtable) on which all the
premises
p
1
, . . . ,
p
n
of an argument are true and the conclusion
q
is
false.
All you need to do here
is (
i
) write down the relevant row of the
truthtable, and (
ii
) say “Here is an interpretation on which
p
1
, . . . ,
p
n
are all true and
q
is false. So,
p
1
, . . . ,
p
n
∴
q
is
in
valid.”
2. You discover that there is
no possible way
of making
p
1
, . . . ,
p
n
true
and
q
false. Here, you need to
explain all of your reasoning
(as I do in
lecture, or as Forbes does, or as I do in my handout). It must be clear
that you have
exhausted all possible cases
, before concluding that
p
1
,
. . . ,
p
n
∴
q
is
valid
. This can be rather involved, and should be spelled
out in a stepbystep fashion. Each salient case has to be examined.