Qual Flashcards
True or false:
All statements are reciprocal
False
All Statements are not reciprocal
All statements always imply the contrapositive
This means the converse is a fallacy of All statements (it is also a fallacy of most statements)
Most
Most implies a majority (>50%) greater than half
True or false:
All statements always imply the contrapositive
True
All Statements are not reciprocal
All statements always imply the contrapositive
True or false:
Most statements are reciprocal
False
Most statements are not reciprocal
Most statements don’t have a contrapositive
Most statements can be all
Because Most statements are not reciprocal, like All statements, they can suffer the converse fallacy
True or false:
Most statements are not reciprocal
True
Most statements are not reciprocal
Most statements don’t have a contrapositive
Most statements can be all
Because Most statements are not reciprocal, like All statements, they can suffer the converse fallacy
True or false:
Most people who play Dungeons and Dragons live in their parents’ house, so you can infer that most people who live in their parents’ house play Dungeons and Dragons
False
False
Most statements are not reciprocal
Most statements don’t have a contrapositive
Most statements can be all
Because Most statements are not reciprocal, like All statements, they can suffer the converse fallacy
DD -m-> PH
A-m-> B does not translate to B-m->A
Diagram most A’s are B’s
A -most-> B
Most vs The Most
Majority vs Plurality
True or false:
Most could be All
True
Most can be All
True or false:
Most cannot be All
False
Most can be all
If most A’s are B’s does it mean that there are A’s that are not B’s
Not necessarily,
Most statements can be more than half or even All
True or false:
Most people who play Dungeons and Dragons live in their parents’ house, therefore, there are some people who play DD who don’t live in their parents’ house
False, we cannot say this for certain
Most statements can be more than half or even All
DD-most->PB does not translate to DD -s->not Pb
A-most->B does not mean A-some-> not B
True or false:
Some statements are reciprocal
True
Some statements are reciprocal
If some A are B’s, then some B’s are A’s
A-some-B is the same as B-some-A
Some statements do not have a contrapositive
Some could be just one
Some could be Most or even All
True or false:
If some A are B’s, then some B’s are A’s
True
Some statements are reciprocal
If some A are B’s, then some B’s are A’s
A-some-B is the same as B-some-A
Some statements do not have a contrapositive
Some could be just one
Some could be Most or even All
True or false:
Some of the people are going.
Therefore, all the people might be going.
True
Some could be just one
Some could be Most or even All
Some statements are reciprocal
If some A are B’s, then some B’s are A’s
A-some-B is the same as B-some-A
Some statements do not have a contrapositive
Easy way of remembering what some translates to
At least one
Some could be just one
Some could be Most or even All
Some statements are reciprocal
If some A are B’s, then some B’s are A’s
A-some-B is the same as B-some-A
Some statements do not have a contrapositive
True or false.
Some A are B’s, therefore, there are some A’s that are not B’s
False. The statement is invalid.
Some could be just one
Some could be Most or even All
A-some-B cannot be understood to be A-some-not B
Some statements are reciprocal
If some A are B’s, then some B’s are A’s
A-some-B is the same as B-some-A
Some statements do not have a contrapositive
True or false:
All A’s are B’s, and some B’s are C’s.
Therefore, Some A’s are C’s
False A->B B-some-C ------------- A-some-C
Invalid^
.
.
A->B
A-some-C
————-
B-some-C
^Valid
.
.
To validly draw an inference from an All statement and a some statement, the term shared by the two statements must be the sufficient condition of the all statement.
In the case above, A wasn’t present in the some statement
All A’s are B’s, and some A’s are C’s.
Therefore, some B’s are C’s
True
A->B
A-some-C
————-
B-some-C
^Valid . . To validly draw an inference from an All statement and a some statement, the term shared by the two statements must be the sufficient condition of the all statement. . . . .
A->B
B-some-C
————-
A-some-C
Invalid^
In the case above, A wasn’t present in the some statement
All A’s are B’s and some C’s are A’s.
Therefore, some C’s are B’s
True
A->B
A-some-C (reversible)
————-
B-some-C (reversible)
^Valid . . To validly draw an inference from an All statement and a some statement, the term shared by the two statements must be the sufficient condition of the all statement. . . . .
A->B
B-some-C
————-
A-some-C
Invalid^
In the case above, A wasn’t present in the some statement
True or false:
All circus performers have extraordinary talents.
Some people with extraordinary talents can swallow swords. Therefore, some circus performers can swallow swords.
False
Circus performer -> Extraordinary Talent
Extraordinary Talent -some-> Swallow Swords
——————
Circus Performer -some-> Swallow Swords
.
^Invalid!
.
.
To validly draw an inference from an All statement and a some statement, the term shared by the two statements must be the sufficient condition of the all statement.
-
The shared term in the argument above is the necessary of the All statement. It has to be the sufficient, the all “blanks”
True or false:
All circus performers have extraordinary talents.
Some people who can swallow swords perform in the circus. Therefore, some people with extraordinary talents can swallow swords
true. the statement is valid
A->B
A-some-C (reversible)
————-
B-some-C (reversible)
^Valid . . To validly draw an inference from an All statement and a Some statement, the term shared by the two statements must be the sufficient condition of the all statement. . . . .
A->B
B-some-C
————-
A-some-C
Invalid^
In the case above, A wasn’t present in the some statement
To validly draw an inference from an All statement and a Some statement, the term shared by the two statements must be the
sufficient condition of the all statement. . A->B ---this A A-some-C (reversible) ------------- B-some-C (reversible)
True or false:
To validly draw an inference from an All statement and a Some statement, the term shared by the two statements must be the necessary condition of the all statement.
False
it is the sufficient assumption that must appear in both statements to produce a some
A->B
B-some-C
————-
A-some-C
Invalid^ . . . A->B A-some-C (reversible) ------------- B-some-C (reversible)
^Valid
.
Most A’s are B’s
All B’s are C’s
A-m->B B -> C ----------- A -most-> C When the Sufficient condition of the All statement is the Necessary of the Most statement, then you can infer a Most Start with sufficient from Most . . A-m-> B A -> C ---------- B -some- C When the sufficient of the All statement matches with the sufficient of the Most statement, then you can infer a Some
All A’s are C’s
Most A’s are B’s
A -> C A-m->B ------------ C -some-B . When the sufficient of the All statement matches with the sufficient of the Most statement, then you can infer a Some . . . A-m->B B -> C ----------- A -most-> C When the Sufficient condition of the All statement is the Necessary of the Most statement, then you can infer a Most
True or false All A's are B's Most B's are C's --------------- Most A's are C's
false
No valid conclusion can be made from this
In order to make any inference from an All and a Most statement,
The shared condition MUST be on the Sufficient of the All
if it is shared with the Necessary of the Most = Most
with Sufficient of Most = Some
To make a deduction from two most sections..
The two sufficient conditions - the thing before the arrow - must be the same. Then draw a some between the two different remaining conditions
A-most->B
A-most->C
————-
B-some-C
Most 1’s are 2’s
All 2’s are 3’s
Most 1’s are 2’s
All 2’s are 3’s
——–
Most 1’s are 3’s
F –> C
+
C (xIx) D
Valid Inference F --> C C (xIx) D ----------- F --> C D ======== F (xIx) D
This is a valid inference
.
.
.
.
Compare that with D --> G \+ C (xIx) D --------- C (xIx) D --> G Not a valid inference
D –> G
+
C (xIx) D
Can’t produce a double arrow unless matching end is on the necessary
D --> G \+ C (xIx) D --------- C (xIx) D --> G Not a valid inference
-only valid conclusion you can get is that “some G’s are not C’s”
Compare with
.
Valid Inference F --> C C (xIx) D ----------- F --> C (xIx) D ======== F (xIx) D
A -> B
B -some- C
Cant get valid inference
Some MUST match SUFFICIENT condition of All
A -> B
A -some- C
———–
B -some- C
A -> B
A -some- C
Valid inference
Some MUST match SUFFICIENT condition of All
A -> B
A -some- C
———–
B -some- C
When adding together a Some statement with an All, the BLANK condition of the All Must be found in the Some
When adding together a Some statement with an All, the SUFFICIENT condition of the All Must be found in the Some
Valid inference
Some MUST match SUFFICIENT condition of All
A -> B
A -some- C
———–
B -some- C
All A’s are B’s
Some A’s are C’s
Therefore, some B’s are C’s
When adding together a Some statement with an All, the SUFFICIENT condition of the All Must be found in the Some
Valid inference
Some MUST match SUFFICIENT condition of All
A -> B
A -some- C
———–
B -some- C
All A’s are B’s
Some B’s are C’s
Therefore, some A’s are C’s
False - Invalid
Some MUST match SUFFICIENT condition of All
A -> B
A -some- C
———–
B -some- C
All C’s are E’s
Some E’s are T’s
No valid conclusion
Invalid
Some MUST match SUFFICIENT condition of All to produce a some
A -> B
A -some- C
———–
B -some- C