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exam! See Mock Exam (syllabus
... click on this week!)
I. Some old issues:
A. Tables: Memorize!
Remember:
| |
P |
Q |
P&Q |
PvQ |
P>Q |
P=Q |
~P |
| row one: |
T |
T |
T |
T |
T |
T |
F |
| row two: |
T |
F |
F |
T |
F |
F |
F |
| row three: |
F |
T |
F |
T |
T |
F |
T |
| row four: |
F |
F |
F |
F |
T |
T |
T |
B. Unless:
- I said: use 'v' for unless. ("We win unless they score":
WvS and anything logically equivalent. Like ~S>W.)
- More ideas? A better example?
- Me:
Another Example
You get X unless problem P gets in the way = you get X if P doesn't
get in the way:
~P>X
Now we just need a method to see that this is equivalent to both 'PvX'
and '~X>P'
C.The following lists a number of equivalences. We've done most already.
For now, look at triple-bar.
| Equivalent English Forms (Each table element
-- i.e., box -- below gives English forms instances of each of which
can be symbolized by a sentence of any SL form on its right. Please
note that there are many more English forms than can be covered below.) |
Equivalent SL Forms (Each table
element below gives SL sentence-forms to guide in translating English
sentences of forms found on the left. Please note that this is an
incomplete list of possible symbolizations.) |
Example Applications (Each of the
table elements below shows a way to apply the table elements on their
left.) |
|
If P, then Q.
If P, Q.
Provided P, Q.
Were P to hold, Q
would be true.
Should P be true, Q.
P only if Q.
P is a sufficient condition for Q.
P implies Q.
|
P>Q
~Q>~P
|
If there is fire, then there is Oxygen" or "There
is fire only if there is oxygen" may both be symbolized
as 'F>O' or equivalently as '~O>~F'. |
P if Q.
P provided Q.
P is a necessary condition for Q.
|
Q>P
~P>~Q
|
"Water is a necessary condition for life"
or "there's water if there's life" may both be symbolized
as 'L>W' or equivalently as ~W>~L'. |
P if and only if Q.
P just in case Q.
P is necessary and sufficient for Q.
|
P=Q
(P>Q)&(Q>P)
|
"An argument is sound if and only if it is both
valid and has true premises" may be symbolized as either 'S=(V&T)'
or '[S>(V&T)] &[(V&T)>S] |
|
Both P
and Q.
P and Q.
P but Q.
Q and P.
Q
but P.
P however Q.
P although Q.
P moreover Q.
|
P&Q
Q&P
|
"Sandra is both brave and careful", "Sandra
is brave, moreover she is careful' or "Sandra is brave but careful"
can all be symbolized as 'B&C' or 'C&B'. |
Either
P or Q.
Either Q or P.
P or Q.
Q or P.
At least one of P, Q.
|
PvQ
QvP |
"Either the other team will score and
tie up the game, or we win!" can be symbolized as '(S&T)vW'. |
P unless Q.
Q unless P.
Unless P, Q.
Unless Q, P.
|
PvQ
~Q>P
~P>Q
|
"We win unless the other team scores" can
be symbolized as 'WvS'. |
Neither P
nor Q.
Not-P and not-Q.
|
~(PvQ)
~P&~Q
|
"They neither scored not tied the game"
may be symbolized as either '~(WvT)' or '~W&~T'. |
It's not the case that both P
and Q.
Not both P and Q.
Either not-P or not-Q.
|
~(P&Q)
~Pv~Q
|
"Sandra is not both brave and careful" may
be symbolized as either '~(B&C)' or '~Bv~C'. |
Harder Symbolizations
(2.4)
You do (Use B,C,L):
- Brazil wins a gold unless Canada does too.
- Brazil wins a gold provided Canada wins a gold.
- Neither Canada nor Luxemburg wins a gold.
- Provided Brazil doesn't win a gold, then both Canada and Luxemburg
win.
- Brazil wins a gold only if neither of the other two win a gold.
- Brazil's winning a gold is a sufficient condition for both Canada
and Luxembrug not winning.
D Informal Proofs
1. Suppose P is logically true, then ~P
is logically false. 1,2,3,4,5
(You do...)
2. If P is logically true, then so is PvQ.
(We do...)
3. Show that PvQ
can be logically true even though both it's components are logically indeterminate.
Hint:
For these problems, be careful to distinguish general proofs from proofs
by example.
- You can give an example as your proof only if this completely justifies
the proposition you are asked to prove.
- When you are asked to prove something about an "arbitrary subject",
then begin by supposing that X is such a subject. Then chase through
definitions.
II. Homework questions?
A. Syntax?
B. Symbolization? More help and problems in Chapter
4. You're ready for that now!
C. Arguments and indicators?
D. Tables and Truth Functionality
III. Tables and section 3.1
A. All our SL connectives are truth functional. That just means that
the truth value of a compound sentence depends on the truth value of the
components.
So, ...
if you know that
P: GWB is prez.
S: GWB is a senator
F: GWB is female
are true, false, false (respectively), then which of the following
are true? false?
-
P&S
False
-
PvS
Rewrite the problem this way:
Then just fill in the '?' with a ...
-
P>S
True
-
(P&S)&F
False
-
P>(S&F)
First, write truth values for the simplest sentences -- the atomic
ones as shown -- then work your way up until you get to the main
connective. Under it, write the final result.
OK, now let's look
at some of the 3.1 problems...and the special tool to help you think
about them.
IV. 3.2's Full Tables
A. What if we don't know t-values? E.g., 'A&~B' about Art and Beth's
being students?
Then...
...we can only think about the different possibilities.
This sentence, (A&~B), would be false if both are students.
Right?
What would make this sentence
true???
Can you guess?
If not then you can go through all the possiblities. a) both are
students, b) he is but she's not, c) he's not but she is, and d) neither
is a student.
But notice that
we've already done
this sort of thing:
We need truth tables again, like so:
B. This is just doing 3.1, but for each way that the atomic sentences
can be true or false.
C. You try: ~A>(B&A)
help?
D. Now, what about 'A&(B&C)'??? How do we get all the possiblities?
Think about what could be true about Barb and Carol, IF Art is
a student...and then IF he is NOT a student. Let's
write this up in a table. Or just look
at the finished product.
| A |
B |
C |
| T |
T |
T |
| T |
T |
F |
| T |
F |
T |
| T |
F |
F |
| F |
T |
T |
| F |
T |
F |
| F |
F |
T |
| F |
F |
F |
If 'A' is true, then notice how 'B' and 'C' can vary.
And, see what happens when 'A' is false.
Now, just do the table...
or just look.
Next Question: How
many rows if we have 4 atomic sentences?
| A |
B |
C |
D |
| T |
T |
T |
T |
| T |
T |
T |
F |
| T |
T |
F |
T |
| T |
T |
F |
F |
| T |
F |
T |
T |
| T |
F |
T |
F |
| T |
F |
F |
T |
| T |
F |
F |
F |
| F |
T |
T |
T |
| F |
T |
T |
F |
| F |
T |
F |
T |
| F |
T |
F |
F |
| F |
F |
T |
T |
| F |
F |
T |
F |
| F |
F |
F |
T |
| F |
F |
F |
F |
This pattern emerges:
| Number of Atomic Components: |
Number of Rows: |
| 1 |
2 |
| 2 |
4 |
| 3 |
8 |
| 4 |
16 |
| n |
2
n
|
Now, we need to do some more. We
do.
And
you do...
1. L&~S
2. (J&K)=(K&J)
3. (L>S)>(S&L)
|