Transcript Practice Quiz 2

Practice Quiz 3
Hurley 4.3 – 4.6
For the quiz …
I will provide you with a categorical proposition, like…
No apples sold in Minnesota are mushy weapons
I’ll ask you for its








quality
qualifier
quantity
quantifier
copula
distribution
letter name
terms
1
Consider:
No non-A are B (T) Obversion
a. Some non-A are B. (F)
b. All A are non-B. (Und.)
c. All non-A are non-B. (T)
d. Some non-A are not B. (T)
e. No B are non-A. (T)
2
Consider:
All A are non-B. (F) Contraposition
a. All A are non-B. (F)
b. All non-B are A. (Und.)
c. No non-A are B. (Und.)
d. All B are non-A. (F)
e. Some non-A are not B. (T)
3
Consider:
Some A are not non-B. (T)  Some A are B.
a.Contraposition (T)
b.Contrary (F)
c.Conversion (T)
d.Obversion (T)
e.Subcontrary (Und.)
4
Consider:
Some non-A are B. (F)  Some B are non-A.
a. Subcontrary (T)
b. Conversion (Und.)
c. Contraposition (Und.)
d. Conversion (F)
e. Contraposition (F)
5
Assume Aristotle (Traditional standpoint).
Consider:
Some A are non-B. (F)  Some A are not non-B. (F)
a. Illicit, contrary
b. Illicit, subalternation
c. Subcontrary
d. Illicit, subcontrary
e. Contraposition
6
No S are P. (Aristotelian standpoint)
After filling in the diagram …
a.Area 2 is shaded, and there is a circled X in area
1.
b.Areas 1 and 3 are shaded.
c.Area 1 is shaded, and there is a circled X in area
2.
d.There is an X in area 2.
e.Area 1 is shaded, and there are no other marks.
7
All S are P. (Boolean standpoint)
After filling in the diagram …
a.Areas 1 and 3 are shaded.
b.Area 2 is shaded, and there are no other marks.
c.Area 1 is shaded, and there is a circled X in area
2.
d.There is an X in area 2.
e.Area 1 is shaded, and there are no other marks.
8
Shade area 2 and place an X in area 1.
Which of the following would be valid inferences:
a.shaded area 2.
b.an X in area 3.
c.an X in area 1.
d.shaded 1.
e.no X’s or shadings.
9
Shade area 1 and place an X in area 2.
Which of the following would be valid inferences:
a.shaded area 2.
b.an X in area 3.
c.shaded area 1, and X in area 2.
d.shaded 1.
e.no X’s or shadings.
10
Assume Aristotle (Traditional standpoint).
Consider:
No non-A are B. (T)  Some non-A are not B. (F)
a. Illicit, subalternation
b. Illicit, contradictory
c. Contradictory
d. Illicit, subcontrary
e. Conversion
11
Assume Boole (Modern standpoint).
Consider:
No A are B. (T)  Some A are B. (F)
a. Existential fallacy
b. Illicit, contradictory
c. Contradictory
d. Illicit, subcontrary
e. Conversion
12
Assume Boole (Modern standpoint).
Consider:
No A are B. (T)  All A are B. (F)
a. Existential fallacy
b. Illicit, contrary
c. Contradictory
d. Illicit, subcontrary
e. Conversion
13
Assume Aristotle (Traditional standpoint)
All square circles are happy shapes.
 Some square circles are happy shapes.
a. Existential fallacy
b. Valid, contradictory
c. Valid, subcontrary
d. Invalid, subalternation
e. Invalid, contrary
Conditional Validity
Remember, if asked to test for conditional validity, assume
Aristotelian standpoint (put a circled X in any unshaded
area of the subject circle of a universal proposition).
Test for validity.
If valid, check to see whether the subject matter of the
proposition exists.
If so,
Conditionally Valid.
If not,
Existential Fallacy.
Example on next slide:
Conditionally Valid?
No Moms are Tired Workers 
Some Moms are not Tired Workers
X
Conditionally Valid, or
Existential Fallacy?
M
T
X
M
T
Answer Key …
I will provide you with a categorical proposition, like…
No apples sold in Minnesota are mushy weapons
I’ll ask you for its








quality (Negative)
qualifier (‘No’)
quantity (Universal)
quantifier (‘No’)
copula (‘are’)
distribution (both Subject and Predicate)
letter name (‘E’)
terms (apples sold in Minnesota=subject term … mushy
weapons=predicate term)
Answer Key
1=C, 2=D, 3=D, 4=D, 5=D, 6=A, 7=E, 8=A and
C, 9=C and D, 10=A, 11=C (valid inference),
12=A, 13=A
Final slide: Conditionally valid because ‘moms’
exist.