**********
* NOTE!!
**********
* we have lecture AS USUAL today and Wednesday, Oct. 10
* BUT, there is NO CS 100 LECTURE *this* Friday, October 12,
because I am travelling to a conference
* (that's why your on-paper answers for HOMEWORK 6 are due
at the beginning of class next MONDAY, October 15)
**********
* also NOTE:
in our 2nd example last time (clean driving record example),
we had our argument rewritten to the form:
P -> Q
P
∴ Q
* does this look familiar?
remember modus ponens:
If A, then B
A
Therefore, B
...in propositional logic form, this is indeed
A -> B
A
∴ B
========
* can we do this with modus TOLLENS?
If A, then B
Not B
Therefore, not A.
YES! in propositional logic, this is written as:
A -> B
~B
∴ ~A
========
* can we do this with the chain argument pattern?
If A, then B
If B, then C
∴ If A, then C
YES! in propositional logic, this is written as:
A -> B
B -> C
∴ A -> C
********
* CONTINUING the "students can raise their grades..." example from
last Friday...
(adapted from course text)
* Students can raise their grade by studying hard,
but not by doing extra-credit work.
* It's not the case that students can raise their
grades by doing extra-credit work and hiring
a tutor.
* SO, students can raise their grade
by studying hard and also by getting a tutor.
* IDENTIFY the buried simple statements,
and assign each a letter!
S = Students can raise their grade by studying hard
E = Students can raise their grade by doing
extra-credit work.
R = Students can raise their grade by getting a tutor.
* rewrite the argument in propositional logic form:
S ^ ~E
~(E ^ R)
∴ S ^ R
* next: we'll build an appropriate truth table!
* start with a column for each letter, and "build" up from there,
1 column per operation until you get all a column for each premise
and the conclusion:
S E R ~E S^~E E^R ~(E^R) S^R
* I recommend, right now, before you forget,
LABEL these with (*) and (C) to mark your premises and conclusion
(*) (*) (C)
S E R ~E S^~E E^R ~(E^R) S^R
* fill in the letter-columns so that all possible combinations of T and F
are covered...
(3 letters, 2 ^ 3rd power combos, 2 * 2 * 2 = 8 rows)
(*) (*) (C)
S E R ~E S^~E E^R ~(E^R) S^R
--------------------------------------
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F
* NOW fill in the rest of the columns, one at a time,
based on the values of columns to the left of it (as appropriate)
(*) (*) (C)
S E R ~E S^~E E^R ~(E^R) S^R
--------------------------------------
T T T F F T F T
T T F F F F T F
T F T T T F T T
T F F T T F T F
F T T F F T F F
F T F F F F T F
F F T T F F T F
F F F T F F T F
* NOW cross out ALL rows for which ANY premise
is False (done on document camera copy)
AND see: in the remaining rows, is the
Conclusion ALWAYS T?
It IS NOT, here, so this argument is INVALID
********
* ONE MORE EXAMPLE:
* If the Green Party loses in the local election, then the
developer-friendly politicians will have a majority in the
city council.
* If the developer-friendly politicians have a majority in the
city council, then the city council will vote down
restrictions on development on agricultural land.
* It is NOT the case that the city council will vote down
restrictions on development on agricultural land OR
that the Green Party will lose in the local election.
* THEREFORE, it is NOT the case that IF the Green Party does NOT
lose in the local election, THEN the city council will NOT
vote down restrictions on development of agricultural land.
* IDENTIFY the simple statements in the argument, and give each
a letter
P = the Green Party loses in the local election
Q = the developer-friendly politicians will have a majority in the
city council
R = the city council will vote down
restrictions on development on agricultural land
* now rewrite the argument in prop logic form:
P -> Q
Q -> R
~(R v P)
∴ ~( ~P -> ~R )
...continuing THIS on Wednesday!