Propositional Equivalences CMSC 56 | Discrete Mathematical Structure for Computer Science August 23, 2018 Instructor: Allyn Joy D. Calcaben College of Arts & Sciences University of the Philippines Visayas

1. Propositional Equivalences
Lecture 2, CMSC 56
Allyn Joy D. Calcaben

2. Definition 1
A compound proposition that is always true, no matter what the
truth values of the propositional variables that occur in it, is called a
tautology.
A compound proposition that is always false is called a
contradiction.
A compound proposition that is neither a tautology nor a
contradiction is called a contingency.
Propositions

3. Truth Table for p V ￢p
Example
p ￢p p V ￢p
T F
T F
F T
F T

4. Truth Table for p V ￢p
Example
p ￢p p V ￢p
T F T
T F T
F T T
F T T

5. Truth Table for p V ￢p
Therefore, p V ￢p is a TAUTOLOGY.
Example
p ￢p p V ￢p
T F T
T F T
F T T
F T T

6. Truth Table for p ∧ ￢p
Example
p ￢p p ∧ ￢p
T F
T F
F T
F T

7. Truth Table for p ∧ ￢p
Example
p ￢p p ∧ ￢p
T F F
T F F
F T F
F T F

8. Truth Table for p ∧ ￢p
Therefore, p ∧ ￢p is a CONTRADICTION.
Example
p ￢p p ∧ ￢p
T F F
T F F
F T F
F T F

9. Logical Equivalences

10. Logical Equivalences
Compound propositions that have the same truth values in all
possible cases are called logically equivalent.

11. Definition 2
The compound propositions p and q are called logically
equivalent if p ↔ q is a tautology.
The notation p ≡ q denotes that p and q are logically
equivalent.
Propositions

12. Note
The symbol ≡ is not a logical connective, and p ≡ q is not a
compound proposition but rather is the statement that p ↔ q
is a tautology.
The symbol ⇔ is sometimes used instead of ≡ to denote
logical equivalence.

13. How would you know if two
compound propositions are
equivalent?

14. How would you know if two
compound propositions are
equivalent?
Ans: TRUTH TABLES

15. Logical Equivalence
The compound propositions p and q are equivalent if and only
if the columns giving their truth values agree.

16. De Morgan’s Laws
De Morgan’s Laws
￢(p ∧ q) ≡ ￢p ∨ ￢q
￢(p ∨ q) ≡ ￢p ∧ ￢q

17. Example
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.

18. Example
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T
T F
F T
F F

19. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T
T F T
F T T
F F F

20. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T F
T F T F
F T T F
F F F T

21. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T F F
T F T F F
F T T F T
F F F T T

22. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T F F F
T F T F F T
F T T F T F
F F F T T T

23. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T F F F F
T F T F F T F
F T T F T F F
F F F T T T T

24. Solution
Show that ￢(p ∨ q) and ￢p ∧￢q are logically equivalent.
p q p ∨ q ￢ ( p V q ) ￢ p ￢ q ￢ p ∧ ￢ q
T T T F F F F
T F T F F T F
F T T F T F F
F F F T T T T

25. Conclusion
Because the truth values of the compound propositions ￢(p
∨ q) and ￢p ∧￢q agree for all possible combinations of the
truth values of p and q, it follows that￢(p ∨ q) ↔ (￢p ∧￢q)
is a tautology and that these compound propositions are
logically equivalent.

26. Solution
Show that ￢(p ∨ q) ↔ (￢p ∧￢q) is a tautology.
￢ ( p V q ) ￢ p ∧ ￢ q ￢(p ∨ q) ↔ (￢p ∧￢q)
F F
F F
F F
T T

27. Solution
Show that ￢(p ∨ q) ↔ (￢p ∧￢q) is a tautology.
￢ ( p V q ) ￢ p ∧ ￢ q ￢(p ∨ q) ↔ (￢p ∧￢q)
F F T
F F T
F F T
T T T

28. Example
Show that p → q and ￢p ∨ q are logically equivalent.

29. Solution
Show that p → q and ￢p ∨ q are logically equivalent.
p q p → q ￢ p ￢ p V q
T T
T F
F T
F F

30. Solution
Show that p → q and ￢p ∨ q are logically equivalent.
p q p → q ￢ p ￢ p V q
T T T
T F F
F T T
F F T

31. Solution
Show that p → q and ￢p ∨ q are logically equivalent.
p q p → q ￢ p ￢ p V q
T T T F
T F F F
F T T T
F F T T

32. Solution
Show that p → q and ￢p ∨ q are logically equivalent.
p q p → q ￢ p ￢ p V q
T T T F T
T F F F F
F T T T T
F F T T T

33. Solution
Show that p → q and ￢p ∨ q are logically equivalent.
p q p → q ￢ p ￢ p V q
T T T F T
T F F F F
F T T T T
F F T T T

34. Example
Show that p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r) are logically
equivalent. This is the distributive law of disjunction over
conjunction.

35. Solution
p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r )
p q r q ∧ r p ∨ (q ∧ r) p ∨ q p ∨ r (p ∨ q) ∧ (p ∨ 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

36. Solution
p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r )
p q r q ∧ r p ∨ (q ∧ r) p ∨ q p ∨ r (p ∨ q) ∧ (p ∨ r)
T T T T T T T T
T T F F T T T T
T F T F T T T T
T F F F T T T T
F T T T T T T T
F T F F F T F F
F F T F F F T F
F F F F F F F F

37. Solution
p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r )
p q r q ∧ r p ∨ (q ∧ r) p ∨ q p ∨ r (p ∨ q) ∧ (p ∨ r)
T T T T T T T T
T T F F T T T T
T F T F T T T T
T F F F T T T T
F T T T T T T T
F T F F F T F F
F F T F F F T F
F F F F F F F F

38. using De Morgan’s Laws
De Morgan’s Laws
￢(p ∧ q) ≡ ￢p ∨ ￢q
￢(p ∨ q) ≡ ￢p ∧ ￢q

39. Example
Use De Morgan’s laws to express the negations of
“Miguel has a cellphone and he has a laptop computer”
and
“Heather will go to the concert or Steve will go to the
concert.”

40. Solution
“Miguel has a cellphone and he has a laptop computer”

41. Solution
“Miguel has a cellphone and he has a laptop computer”
￢(p ∧ q)

42. Solution
“Miguel has a cellphone and he has a laptop computer”
￢(p ∧ q)

43. Solution
“Miguel has a cellphone and he has a laptop computer”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)

44. Solution
“Miguel has a cellphone and he has a laptop computer”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)
The negation is therefore

45. Solution
“Miguel has a cellphone and he has a laptop computer”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)
The negation is therefore
“Miguel does not have a cellphone or he does not have a
laptop computer.”

46. Solution
“Heather will go to the concert or Steve will go to the
concert.”

47. Solution
“Heather will go to the concert or Steve will go to the
concert.”
￢(p ∧ q)

48. Solution
“Heather will go to the concert or Steve will go to the
concert.”
￢(p ∧ q)

49. Solution
“Heather will go to the concert or Steve will go to the
concert.”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)

50. Solution
“Heather will go to the concert or Steve will go to the
concert.”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)
The negation is therefore

51. Solution
“Heather will go to the concert or Steve will go to the
concert.”
￢(p ∧ q) = ￢p ∨￢q (De Morgan’s Laws)
The negation is therefore
“Heather will not go to the concert and Steve will not go to
the concert.”

52. Constructing New Logical
Equivalences

53. Logical Equivalences
Equivalence Name
p ∧ T ≡ p
p ∨ F ≡ p
Identity Laws
p ∨ T ≡ T
p ∧ F ≡ F
Domination Laws
p ∨ p ≡ p
p ∧ p ≡ p
Idempotent Laws
￢(￢p) ≡ p Double Negation Laws
p ∨ q ≡ q ∨ p
p ∧ q ≡ q ∧ p
Commutative Laws

54. Logical Equivalences
Equivalence Name
(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
(p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
Associative Laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Distributive Laws
￢(p ∧ q) ≡ ￢p ∨￢q
￢(p ∨ q) ≡ ￢p ∧￢q
De Morgan’s Laws
p ∨ (p ∧ q) ≡ p
p ∧ (p ∨ q) ≡ p
Absorption Laws
p ∨￢p ≡ T
p ∧￢p ≡ F
Negation Laws

55. Logical Equivalences
Equivalence Name
p ∨ ￢p ≡ T Tautology
p ∧ ￢p ≡ F Contradiction
p → q ≡ ￢p ∨ q Implication Equivalence
p ↔ q ≡ (p → q) ∧ (q → p) Biconditional Equivalence

56. Example
Show that ￢(p → q) and p ∧￢q are logically equivalent.
Use logical identities that we already know to establish new
logical identities.

57. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example

58. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ by the second De Morgan law

59. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ ￢(￢p)∧￢q by the second De Morgan law
second De Morgan law: ￢(p ∨ q) ≡ ￢p ∧￢q

60. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ ￢(￢p)∧￢q by the second De Morgan law

61. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ ￢(￢p)∧￢q by the second De Morgan law
≡ by the double negation law

62. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ ￢(￢p)∧￢q by the second De Morgan law
≡ by the double negation law
double negation law: ￢(￢p) ≡ p

63. Solution
Starting with ￢(p → q) and ending with p ∧￢q, we have the
following equivalences.
￢(p → q) ≡ ￢(￢p ∨ q) by previous example
≡ ￢(￢p)∧￢q by the second De Morgan law
≡ p ∧￢q by the double negation law

64. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ by the second De Morgan law
≡ by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

65. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ by the second De Morgan law
≡ by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
second De Morgan law: ￢(p ∨ q) ≡ ￢p ∧￢q

66. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

67. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
first De Morgan law: ￢(p ∧ q) ≡ ￢p ∨￢q

68. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

69. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
double negation law: ￢(￢p) ≡ p

70. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

71. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
second distributive law: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

72. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

73. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
Contradiction: p ∧ ￢p ≡ F

74. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ F ∨ (￢p ∧￢q) by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F

75. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ F ∨ (￢p ∧￢q) by contradiction
≡ by the commutative law for disjunction
≡ by the identity law for F
mutative law for disjunction: p ∨ q ≡ q ∨ p

76. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ F ∨ (￢p ∧￢q) by contradiction
≡ (￢p ∧￢q) ∨ F by the commutative law for disjunction
≡ by the identity law for F

77. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ F ∨ (￢p ∧￢q) by contradiction
≡ (￢p ∧￢q) ∨ F by the commutative law for disjunction
≡ by the identity law for F
the identity law for F: p ∨ F ≡ p

78. Answer
Show that ￢(p ∨ (￢p ∧ q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
￢(p ∨ (￢p ∧ q)) ≡ ￢p ∧￢(￢p ∧ q) by the second De Morgan law
≡ ￢p ∧ [￢(￢p)∨￢q] by the first De Morgan law
≡ ￢p ∧ (p ∨￢q) by the double negation law
≡ (￢p ∧ p) ∨ (￢p ∧￢q) by the second distributive law
≡ F ∨ (￢p ∧￢q) by contradiction
≡ (￢p ∧￢q) ∨ F by the commutative law for disjunction
≡ ￢p ∧￢q by the identity law for F

79. Equivalence Name
p ∧ T ≡ p
p ∨ F ≡ p
Identity Laws
p ∨ T ≡ T
p ∧ F ≡ F
Domination Laws
p ∨ p ≡ p
p ∧ p ≡ p
Idempotent Laws
￢(￢p) ≡ p
Double Negation
Laws
p ∨ q ≡ q ∨ p
p ∧ q ≡ q ∧ p
Commutative Laws
Equivalence Name
(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
(p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
Associative Laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Distributive Laws
￢(p ∧ q) ≡ ￢p ∨￢q
￢(p ∨ q) ≡ ￢p ∧￢q
De Morgan’s Laws
p ∨ (p ∧ q) ≡ p
p ∧ (p ∨ q) ≡ p
Absorption Laws
p ∨￢p ≡ T
p ∧￢p ≡ F
Negation Laws
Equivalence Name
p ∨ ￢p ≡ T Tautology
p ∧ ￢p ≡ F Contradiction
p → q ≡ ￢p ∨ q Implication Equivalence
p ↔ q ≡ (p → q) ∧ (q → p) Biconditional Equivalence

80. Reference
CHAPTER 1.3 - Rosen, Kenneth H.,Discrete Mathematics and
Its Applications, 7th edition, 2012.

81. Use truth tables to verify the commutative laws
1) p ∨ q ≡ q ∨ p.
2) 2) p ∧ q ≡ q ∧ p.
3) Use a truth table to verify the distributive law: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Use De Morgan’s laws to find the negation of each of the following statements.
4) Jan is rich and happy.
5) Carlos will bicycle or run tomorrow.
6) Mei walks or takes the bus to class.
7) Ibrahim is smart and hard working.
Show that each of these conditional statements is a tautology by using truth tables.
8) (p ∧ q) → p 11) p → (p ∨ q)
9) ￢p → (p → q) 12) (p ∧ q) → (p → q)
10) ￢(p → q) → p 13) ￢(p → q)→￢q
14) Prove: (p ∧ ￢q) V q ⇔ p V q using logical equivalences

82. 14) Prove:
(p ∧ ￢q) V q ⇔ p V q
using logical equivalences
Equivalence Name
p ∧ T ≡ p
p ∨ F ≡ p
Identity Laws
p ∨ T ≡ T
p ∧ F ≡ F
Domination Laws
p ∨ p ≡ p
p ∧ p ≡ p
Idempotent Laws
￢(￢p) ≡ p
Double Negation
Laws
p ∨ q ≡ q ∨ p
p ∧ q ≡ q ∧ p
Commutative Laws
Equivalence Name
(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
(p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
Associative Laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Distributive Laws
￢(p ∧ q) ≡ ￢p ∨￢q
￢(p ∨ q) ≡ ￢p ∧￢q
De Morgan’s Laws
p ∨ (p ∧ q) ≡ p
p ∧ (p ∨ q) ≡ p
Absorption Laws
p ∨￢p ≡ T
p ∧￢p ≡ F
Negation Laws
Equivalence Name
p ∨ ￢p ≡ T Tautology
p ∧ ￢p ≡ F Contradiction
p → q ≡ ￢p ∨ q Implication Equivalence
p ↔ q ≡ (p → q) ∧ (q → p) Biconditional Equivalence