Apply the Law of Detachment and the Law of Syllogism in logical reasoning. Set up simple proofs

1. Syllogism and Proofs
Objectives
The student is able to (I can):
• Apply the Law of Detachment and the Law of Syllogism in
logical reasoning
• Set up simple proofs

2. Example
Law of Detachment
• If p → q is a true statement and p is
true, then q is true.
Determine if the conjecture is valid by the
Law of Detachment.
Given: If a student passes his classes, the
student is eligible to play sports.
Ramon passed his classes.
Conjecture: Ramon is eligible to play
sports. validvalidvalidvalid

3. Example Determine if the conjecture is valid by the
Law of Detachment.
Given: If you are tardy 3 times, you must
go to detention. Sheyla is in
detention.
Conjecture: Sheyla was tardy at least 3
times. not validnot validnot validnot valid

4. Examples
Law of Syllogism
• If p → q and q → r are true statements,
then p → r is a true statement.
Determine if each conjecture is valid by the
Law of Syllogism.
Given: If a number is divisible by 4, then it
is divisible by 2. If a number is even,
then it is divisible by 2.
Conjecture: If a number is divisible by 4,
then it is even.
p: A number is divisible by 4
q: A number is divisible by 2
r: A number is even
p → q and r → q; therefore, p → r
not validnot validnot validnot valid

5. Determine if each conjecture is valid by the
Law of Syllogism.
Given: If an animal is a mammal, then it
has hair. If an animal is a dog, then
it is a mammal.
Conjecture: If an animal is a dog, then it
has hair.
p: An animal is a mammal
q: It has hair
r: An animal is a dog
p → q and r → p, therefore r → q
or r → p and, p → q therefore r → q
validvalidvalidvalid

6. Example
We can also use syllogisms to set up chains
of conditionals.
What can we conclude from the following
chain?
If you study hard, then you will earn a good
grade. (p → q)
If you earn a good grade, then your family
will be happy. (q → r)
Conclusion: If you study hard, then your
family will be happy. (p → r)

7. Write a concluding statement:
a → b
d → ~c
~c → a
b → f

8. Write a concluding statement:
a → b d → ~c
d → ~c ~c → a
~c → a a → b
b → f b → f
Conc.: d → f

9. proof An argument, a justification, or a reason
that something is true. To write a proof,
you must be able to justify statements
using properties, postulates, or definitions.
Example: Name the property, postulate, or
definition that justifies each statement.
StatementStatementStatementStatement JustificationJustificationJustificationJustification
If ∠A is a right angle, then
m∠A = 90°.
Definition of right angle
If ∠2 ≅ ∠1 and ∠1 ≅ ∠5,
then ∠2 ≅ ∠5
Transitive property
m∠ABD+m∠DBC=m∠ABC Angle Addition Post.
If B is the midpoint of ,
then AB = BC.
Definition of midpointAC

10. In a lot of ways, proofs are just expanded syllogisms. We are
still setting up chains of statements; the main difference is
that we also have to provide justifications.
Consider the following:
If 3x + 2 = x + 14, then 2x + 2 = 14. (subtraction prop.)
If 2x + 2 = 14, then 2x = 12. (subtraction prop.)
If 2x = 12, then x = 6. (division prop.)
Now look at this as a proof:
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. 3x + 2 = x + 14 1. Given statement
2. 2x + 2 = 14 2. Subtr. prop.
3. 2x = 12 3. Subtr. prop.
4. x = 6 4. Division prop.

11. properties of
congruence
Line segments with equal lengths are
congruent, and angles with equal measures
are also congruent. Therefore, the reflexive,
symmetric, and transitive properties of
equality have corresponding properties ofproperties ofproperties ofproperties of
congruencecongruencecongruencecongruence.
Reflexive Property of Congruence
fig. A ≅ fig. A
Symmetric Property of Congruence
If fig. A ≅ fig. B, then fig. B ≅ fig. A.
Transitive Property of Congruence
If fig. A ≅ fig. B and fig. B ≅ fig. C,
then fig. A ≅ fig. C.

12. Example:
Given:Given:Given:Given: AM bisects
Prove:Prove:Prove:Prove:
Note: While the first reason is almost always “Given”, the
last reason is nevernevernevernever “Prove”. In fact, “Prove” is never evernever evernever evernever ever
used as a reason in a proof (everevereverever)....
Note #2: The last statement of your proof should always be
what you are trying to prove.
CD
CM MD≅
•
D
M
C
•
A
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. bisects 1. Given
2. 2. Def. of midpoint
AM CD
CD MD≅

13. Example:
Given:Given:Given:Given: p ⊥ r
Prove:Prove:Prove:Prove: ∠1 and ∠2 are complementary
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. p ⊥ r 1. Given
2. m∠PAT = 90° 2. Def. of ⊥
3. m∠PAT = m∠1 + m∠2 3. Angle Add. Post.
4. m∠1 + m∠2 = 90° 4. Substitution prop.
5. ∠1 and ∠2 are comp. 5. Def. of comp. ∠s
p
r
1
2
•
S
•
T
•
A
•P

14. Example:
Given:Given:Given:Given: NA = LE
N is midpt of
L is midpt of
Prove:Prove:Prove:Prove:
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. NA = LE
N is midpt of
L is midpt of
1. Given
2. 2. Def. of midpoint
3. 3. Def. of ≅ segments
4. 4. Substitution prop.
5. 5. Substitution prop.
•
E
L
G
N
A
GA
GE
GN GL≅
GA
GE
GN NA; GL LE≅ ≅
NA LE≅
GN LE≅
GN GL≅

15. GivenGivenGivenGiven:::: ∠BAC is a right angle;
∠2 ≅ ∠3
ProveProveProveProve:::: ∠1 and ∠3 are complementary
1
2
3
•
•
B
A C
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. ∠BAC is a right angle 1. Given
2. m∠BAC = 90° 2. _______________
3. __________________ 3. Angle Add. post.
4. m∠1 + m∠2 = 90° 4. Subst. prop.
5. ∠2 ≅ ∠3 5. Given
6. __________________ 6. Def. ≅ ∠s
7. m∠1 + m∠3 = 90° 7. _______________
8. __________________ 8. Def. comp. ∠s

16. GivenGivenGivenGiven:::: ∠BAC is a right angle;
∠2 ≅ ∠3
ProveProveProveProve:::: ∠1 and ∠3 are complementary
1
2
3
•
•
B
A C
StatementsStatementsStatementsStatements ReasonsReasonsReasonsReasons
1. ∠BAC is a right angle 1. Given
2. m∠BAC = 90° 2. _______________
3. ____________________ 3. Angle Add. post.
4. m∠1 + m∠2 = 90° 4. Subst. prop. =
5. ∠2 ≅ ∠3 5. Given
6. ____________________ 6. Def. ≅ ∠s
7. m∠1 + m∠3 = 90° 7. _______________
8. ____________________ 8. Def. comp. ∠s
Def. rightDef. rightDef. rightDef. right ∠∠∠∠
mmmm∠∠∠∠1 + m1 + m1 + m1 + m∠∠∠∠2 = m2 = m2 = m2 = m∠∠∠∠BACBACBACBAC
Subst. prop. =Subst. prop. =Subst. prop. =Subst. prop. =
mmmm∠∠∠∠2 = m2 = m2 = m2 = m∠∠∠∠3333
∠∠∠∠1 and1 and1 and1 and ∠∠∠∠3 are comp.3 are comp.3 are comp.3 are comp.