1. Propositional Equivalences
CS/APMA 202, Spring 2005
Rosen, section 1.2
Aaron Bloomfield

2. Tautology and Contradiction
A tautology is a statement that is always true
p  ¬p will always be true (Negation Law)
A contradiction is a statement that is always false
p  ¬p will always be false (Negation Law) p
p  ¬p
p  ¬p T F

3. Logical Equivalence
A logical equivalence means that the two sides always have the same truth values
Symbol is ≡or  (we’ll use ≡)

4. Logical Equivalences of And
p  T ≡ p Identity law
p  F ≡ F Domination law p T
pT F p F
pF T

5. Logical Equivalences of And
p  p ≡ p Idempotent law
p  q ≡ q  p Commutative law p
pp T F p q
pq
qp T F

6. Logical Equivalences of And
(p  q)  r ≡ p  (q  r) Associative law p q r
pq
(pq)r
qr
p(qr) T F

7. Logical Equivalences of Or
p  T ≡ T Identity law
p  F ≡ p Domination law
p  p ≡ p Idempotent law
p  q ≡ q  p Commutative law
(p  q)  r ≡ p  (q  r) Associative law

8. Corollary of the Associative Law
(p  q)  r ≡ p  q  r
(p  q)  r ≡ p  q  r
Similar to (3+4)+5 = 3+4+5
Only works if ALL the operators are the same!

9. Logical Equivalences of Not
¬(¬p) ≡ p Double negation law
p  ¬p ≡ T Negation law
p  ¬p ≡ F Negation law

10. Sidewalk chalk guy
Source:

11. DeMorgan’s Law Probably the most important logical equivalence
To negate pq (or pq), you “flip” the sign, and negate BOTH p and q
Thus, ¬(p  q) ≡ ¬p  ¬q
Thus, ¬(p  q) ≡ ¬p  ¬q p q p q
pq
(pq)
pq
pq
(pq)
pq T F

12. Yet more equivalences Distributive: Absorption
p  (q  r) ≡ (p  q)  (p  r)
p  (q  r) ≡ (p  q)  (p  r)
Absorption
p  (p  q) ≡ p
p  (p  q) ≡ p

13. How to prove two propositions are equivalent?
Two methods:
Using truth tables
Not good for long formula
In this course, only allowed if specifically stated!
Using the logical equivalences
The preferred method
Example: Rosen question 23, page 27
Show that:

14. Using Truth Tables (pq) →r pq (p→r)(q →r) q →r p→r r q p p q r p→rF

15. Using Logical Equivalences
Original statement
Definition of implication
DeMorgan’s Law
Associativity of Or
Re-arranging
Idempotent Law

16. Quick survey I understood the logical equivalences on the last slide
Very well
Okay
Not really
Not at all

17. Logical Thinking At a trial:
Bill says: “Sue is guilty and Fred is innocent.”
Sue says: “If Bill is guilty, then so is Fred.”
Fred says: “I am innocent, but at least one of the others is guilty.”
Let b = Bill is innocent, f = Fred is innocent, and s = Sue is innocent
Statements are:
¬s  f
¬b → ¬f
f  (¬b  ¬s)

18. Can all of their statements be true?
Show: (¬s  f)  (¬b → ¬f)  (f  (¬b  ¬s)) ¬b ¬f ¬s
¬b→¬f
f(¬b¬s)
¬b¬s
¬sf s f b b f s ¬b ¬f ¬s
¬sf
¬b→¬f
¬b¬s
f(¬b¬s) T F

19. Are all of their statements true
Are all of their statements true? Show values for s, b, and f such that the equation is true
Original statement
Definition of implication
Associativity of AND
Re-arranging
Idempotent law
Absorption law
Distributive law
Negation law
Domination law

20. What if it weren’t possible to assign such values to s, b, and f?
Original statement
Definition of implication
... (same as previous slide)
Domination law
Re-arranging
Negation law
Contradiction!

21. Quick survey I feel I can prove a logical equivalence myself
Absolutely
With a bit more practice
Not really
Not at all

22. Logic Puzzles Rosen, page 20, questions 51-55
Knights always tell the truth, knaves always lie
A says “At least one of us is a knave” and B says nothing
A says “The two of us are both knights” and B says “A is a knave”
A says “I am a knave or B is a knight” and B says nothing
Both A and B say “I am a knight”
A says “We are both knaves” and B says nothing

23. Sand Castles

24. Functional completeness
Functional completeness is discussed on page 27 (questions 37-39) of the text
All the “extended” operators have equivalences using only the 3 basic operators (and, or, not)
The extended operators: nand, nor, xor, conditional, bi-conditional
Given a limited set of operators, can you write an equivalence of the 3 basic operators?
If so, then that group of operators is functionally complete

25. Rosen, § 1.2 question 46 Show that | (NAND) is functionally complete
Equivalence of NOT:
p | p ≡ p
(p  p) ≡ p Equivalence of NAND
(p) ≡ p Idempotent law

26. Rosen, § 1.2 question 46 Equivalence of AND: Equivalence of OR:
p  q ≡ (p | q) Definition of nand
p | p How to do a not using nands
(p | q) | (p | q) Negation of (p | q)
Equivalence of OR:
p  q ≡ (p  q) DeMorgan’s equivalence of OR
As we can do AND and OR with NANDs, we can thus do ORs with NANDs
Thus, NAND is functionally complete

27. Quick survey I felt I understood the material in this slide set…
Very well
With some review, I’ll be good
Not really
Not at all

28. Quick survey The pace of the lecture for this slide set was… Fast
About right
A little slow
Too slow