Transcript 2103-390: Objectives

Logic and Reasoning
1
Objective
 Spot valid and invalid reasoning.
 Be able to construct a valid reasoning.
 Make appropriate predictions based on acceptable premises.
 Logically draw conclusions from experimental result.
Statement
VS
Reasoning
Statement – True or False
Reasoning – Valid or Invalid
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Logic and Reasoning
Premise
Premise
(something assumed
to be true)
If you study hard, you will get A.
You study hard.
Reasoning
Reasoning
You will get A
Conclusion
Conclusion
(something derived
from the premises)
Conclusion/Premise: True/False (T/F)
Reasoning:
Valid/Invalid
(V/I)
In math term,
Premise is called Axiom,
Conclusion is called Theorem, Lemma,
Reasoning is called Proof.
“False conclusion may comes from
invalid reasoning or false premises”.
Only all true premises and
valid reasoning canguarantee
true conclusion.
In experimental science,
Empirical scientists tell us whether statements
are true.
Logicians tell us whether reasoning is valid.
Truth VS Validity

Reasoning
They are not the same.
Premises:
Conclusion
Premises
Truth for statements.
Validity for argument/reasoning.
Dogs have eight legs.
[If x is a dog, then x has eight legs.]
Spooky is a dog.
Conclusion:
Spooky has eight legs.
The argument is valid.
However, the conclusion is false.
For further clarification, see lecture note.
pq
p
valid

q
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Completeness and Soundness Theorems
An argument is valid if and only if it is derivable.
Validity:
Truth of premise guarantees truth of conclusion
Derivability:
kinda like we derive fluid,
according to some rules of inference
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Note
Reasoning
Premises
Conclusion
 Valid reasoning does not guarantee a true conclusion.
 Invalid reasoning does not guarantee a false conclusion.
 A false conclusion does not guarantee invalidity.
 True premises and a true conclusion together do not
guarantee validity.
No valid argument can have true premise and false conclusion.
Some Important Equivalent
… from checking the truth table …
(p)

p
pq

q p
pq

q p
( p  q)  r

p  (q  r )
( p  q)  r

p  (q  r )
p  (q  r )

( p  q)  ( p  r )
p  (q  r )

( p  q)  ( p  r )
5.
pq

p  q
6. Contra-positive
pq

q   p
7.
pq

8.
pq

1. Double Negation
2. Commutative Law
3. Associative Law
4. Distributive Law
9. De Morgan’s Law
10.
11.
( p  q)  (q  p)
q p
 ( p  q)

p  q
 ( p  q)

p  q
( p  q)
( p  q)


pq
 p  q
 p  (q )
 (q )  p
 q  p
p  q
( p  q)  (p  q)
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Are these arguments/reasoning valid or invalid?
Argument 1
pq
q
valid?

p
Argument 3
pq
p
valid?

q
Argument 2
pq
p
valid?

q
Argument 4
pq
q
valid?

p
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Argument 1: Are these arguments/reasoning valid or invalid?
Premises:
If it rains, then the garden is wet.
The garden is wet.
Conclusion:
It rains.
Activity:
pq
q
invalid

p
Class Discussion
Ex)
Premises:
If x = 2p, then sin x = 0.
sin x = 0.
Conclusion:
Therefore, x = 2p.
Invalid
e.g., x = p
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Showing one counter-example is enough for confirming invalid reasoning.
Argument 2: Are these arguments/reasoning valid or invalid?
Premises:
If it rains, the garden is wet.
It rains.
Conclusion:
The garden is wet.
Activity:
pq
p
valid

q
(next page)
Class Discussion
Ex)
Premises:
If x = 2p, then sin x = 0.
x = 2p.
Conclusion:
Valid?
Therefore, sin x = 0.
Showing one true examples is not enough for confirming invalid reasoning.
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You need to show that all possible cases are true.
How to investigate validity
of the reasoning (argument)
Truth
Table
pq
p
No valid argument can has
true premise and false conclusion.
q
T
T
T
T
F
F
F
 p  q   p   q
Logic
Derivation
p  q  p  q   p  p  q   p   q
p
Valid?

q
  p  q   p   q
  ( p  p )  ( q  p )   q
 p  q
 (p  q)
T
T
F
F
T

T
T
F
T
F
T
F
T
 q  p  q
 (q  q )  p
 T  p

 F  (q  p)   q
 (q  p)   q
 (q  p)  q
 T
 p  q   p   q is Tautology?
Try to find Counter-Example,
then show the Contradiction
p
T
x
q

 p  q
T
T
T
Contradiction
F
 p   q
T
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Proof of Valid Reasoning
PQ
by Contradiction Method
No valid argument can have true premise and false conclusion.
PQ
is invalid

at least one case that
P  Q


T
[Using Contra-positive Equivalence]
no one case that
P  Q


T

PQ
is valid
F
Proof by Contradiction Method
 Assume that there is one case that
P  Q


T
F
 Then show that this is not possible – there is no such case by (finding) contradiction.
F
Valid Reasoning (Argument) P  Q
 A reasoning (an argument)
PQ
is said to be valid if and only if, by virtue of logic,
 the truth of the premise P guarantees the truth of the conclusion Q,
 if P is true, Q is necessarily/always true,

PQ
is a tautology.
 In this case, we write
PQ
 A reasoning that is not valid is said to be invalid.
Argument 3: Are these arguments/reasoning valid or invalid?
Premises:
If it rains, the garden is wet.
pq
p
invalid

q
It does not rain.
Conclusion:
The garden is not wet.
Activity:
Premises:
If x = 2p, then sin x = 0.
x
Conclusion:
Class Discussion
2p.
Therefore, sin x
0.
Invalid
e.g., x = p, sin x  0
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Argument 4: Are these arguments/reasoning valid or invalid?
Premises:
If it rains, the garden is wet.
pq
q
   valid

p
The garden is not wet.
Conclusion:
It does not rain.
Activity:
Premises:
Class Discussion
If x = 2p, then sin x = 0.
sin x
Conclusion:
(next page)
Valid?
0.
Therefore, x
2p.
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How to investigate validity
of the reasoning (argument)
pq
q
Valid?

p
Try to find Counter-Example, then show the Contradiction
T

Contradiction
F
p
 q
  q  
F
T
Truth
Table

p
T
T
 p  q   q 
 p
p
q
pq
T
T
T
F
F
T
T
F
F
T
F
T
F
T
T
F
F
T
F
F
T
T
T
T
 p  q   q   p is Tautology?
q
x
F
No valid argument can has
true premise and false conclusion.
 q  p  q   q
p
Logic
Derivation
 p  q
 (p  q)
 p  q   q   p
  p  q   q   p
  (p  q )  ( q  q )   p
  (p  q )  F   p
  (p  q )   p
 (p  q )  p
 (p  q )  p
 (p  p )  q
 T  q
 T
How to investigate validity
of the reasoning (argument)
Argument2
(already
proofed)
pq
p valid
pq

q  p
pq 
 qp
q
valid?
p
q


Contra-positive Equivalent
valid
 is q
is p
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Rule of Inference
Modus
Ponendo
Ponens
pq
pq
p valid
Modus
Tollendo
Tollens
q
q
valid
p
Logical Fallacies

Fallacy of The Converse

Fallacy of The Converse
pq
pq
q
p
p
q
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Some Important Implications
1. Modus Ponens
( p  q)  p

q
2. Modus Tollens
( p  q)  q

p
3. Simplification
4. Addition
5. Modus Tollendo Ponens
6. Hypothetical Syllogism
7. Biconditional-Conditional
8. Conditional- Biconditional
9. Constructive dilemma
pq

p
pq

q
p

pq
q

pq
( p  q)  p

q
( p  q)  q

p
( p  q)  (q  p) 
pq
pq

q p
pq

pq
( p  q)  (q  r ) 
pr
( p  q)  (r  s )  ( p  r ) 
qs
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Logically Draw Conclusions
Premises:
She does not like A and she likes B.
She does not like B or she likes U.
If she likes U, then U are happy.
Conclusions:
She likes who?
and
Who are happy?
Activity:
Class Discussion
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A
=
She likes A.
B
=
She likes B.
U
=
She likes U.
H
=
U are happy.
Premises:
She does not like A and she likes B.
A B
She does not like B or she likes U.
BU
If she likes U, then U are happy.
Conclusions:
U H
?
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A  B
……… (1)
B  U
……… (2)
U H
From (1) with Simplification
From (2) and (5)
with Modus Tollendo Ponens
From (3) and (6)
with Modus Ponens
Premises are assumes
to be true.
……… (3)
A
……… (4)
B
……… (5)
She likes B.
U
……… (6)
She likes U.
H
……… (7)
She doesn’t like A.
U are happy.
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Logically Draw Conclusions
Premises:
If it rains or it is humid, then I wear blue shirt.
If it is cold, then I do not wear blue shirt.
It rains.
Conclusions:
What is the weather condition?
What color of the shirt I wear?
Activity:
Class Discussion
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R
=
It rains.
H
=
It is humid.
B
=
I wear blue shirt.
C
=
Premises:
It is cold.
If it rains or it is humid, then I wear blue shirt.
(R  H )  B
If it is cold, then I do not wear blue shirt.
C  B
It rains.
R
Conclusions:
?
Activity:
Class Discussion
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(R  H )  B
……… (1)
C  B
……… (2)
R
It rains
Premises are assumed
to be true.
……… (3)
From (3) with addition
From (1),(4) with Modus Ponens
From (2),(5) with Modus Tollens
RH
……… (4)
B
……… (5)
C
……… (6)
I wear blue shirt.
It is not cold.
However, we can’t determine the truth value of H.
(we don’t know whether it is humid or not.
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Logically Draw Conclusions
Premises:
If I am bored, then I go to a movie.
If I am not bored, then I go to a library.
If I do not go to a movie, then I do not go to a library.
Conclusions: Where do I go?
Activity:
Class Discussion
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B
=
I am bored.
M
=
I go to a movie.
L
=
I go to a library.
Premises:
If I am not bored, then I go to a library.
B M
B  L
If I do not go to a movie, then I do not go to a library.
M  L
If I am bored, then I go to a movie.
Conclusions:
?
Activity:
Class Discussion
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B M
B  L
M  L
……… (1)
……… (3)
From (3) with Contrapositive
From (2),(4) with
Hypothetical Syllogism
By Tertium non datur
(Principle of Excluded Middle)
From (1),(5),(6) with
Constructive dilemma
Premises are assumed
to be true.
……… (2)
LM
……… (4)
B  M
……… (5)
B  B
M
……… (6)
……… (7)
I goes to a movie.
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If-then Example
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Necessary and Sufficient Condition
Example) The one who graduates, must pass this course.
P: Graduator,
Q: the one who passes this course.
pq
What is the relation
between P and Q?
p q
?
p ? q
p q
?
q ? p
p q
?
x
p
 q  p
x
q
P is necessary or sufficient condition of Q ?
P is sufficient condition of Q.
Q is necessary or sufficient condition of P ?
Q is necessary condition of P.
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Example of statement usually used in conversation
Example)
P only if Q
“จะเป็ น p ได้ ตองเป็
น q
้
(เท่านั้น)”
q p
p
q
pq
แต่การที่เป็ น q ไม่ได้แปลว่าจะเป็ น p
โดยอัตโนมัติ
การที่ไม่ใช่ q นั้น แสดงว่าไม่ใช่ p
q  p
What is necessary / sufficient condition of what ?
P is sufficient condition of Q.
Q is necessary condition of P.
Example of statement usually used in conversation
Example)
q
P if and only if Q
q p
P if Q
p
P only if Q
pq
q
p
pq
Some Logic: Necessary and Sufficient Conditions (Deductive Reasoning)

Implication (Conditional Statement): p  q
p
p
p
q
pq
(~q)  (~p)
Equivalence
If p, then q.
q if p.
q
q
If not q, then not p.
q whenever p
p only if q.
q is a necessary condition for p.
If not q, then not p.
p is a sufficient condition for q.
If p, then q.
Converse of p  q:
qp
Contra-positive of p  q:
~q~p
 p  q and its contra-positive ~ q  ~ p are equivalent. That is:
 If p  q is true, ~q  ~ p is also true.
 If p  q is false, ~ q  ~ p is also false.
 On the other hand, p  q does not imply q  p.
 The truth of p  q does not automatically guarantee the truth of q  p.

Note: There is also “inductive reasoning.”
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Conditional Statements: If p, then q:
PV = mRT
PV  mRT
(If/Under-the-condition-of/) For a fixed gas and mass of the gas,
(if/under-the-condition-of/) and for a fixed temperature:
If volume increases, then pressure decreases.
Vi  Pd
If pressure does not decrease, then volume does not increase.
Pd  Vi
Pressure decreases or volume does not increase.
Vi  Pd
For a fixed gas and mass of the gas, and for a fixed pressure:
If temperature decreases, then volume decreases.
If volume does not decreases, then temperature doest not decreases.
Temperature does not decreases, or volume decreases.
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Real Life Example
Objective:
yB  f ( P ; ... ; ...)
by experiment
varying P
y
By the way,
the experimental result
should be ….?
Premises
Design Exp:
…..
Doing Exp:
…..
Result:
P
Basic Knowledge
Of Mech Material
yB
conclusion
Predicted Result
lab
11 PL3
yB  
2912 EI
P
P
theory
theory
Discussion:
lab
yB
yB
inaccurate E
Not likely possible
maybe possible
Eused ? Ereal
inaccurate I
Not likely possible
maybe possible
Iused ? Ireal
inaccurate L
Not likely possible
maybe possible
Lused ? Lreal
Cause of Error?
lab
Rule of Inference
Modus
Ponendo
Ponens
pq
pq
p valid
Modus
Tollendo
Tollens
q
pq

q
valid
p
q  p
- Investigate the validity of argument (reasoning).
- Make a theoretical predictions. //
Logically draw conclusions.
- Hypothesis Testing
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