Proofs Using Logical
Equivalences
Program: BS ECE (LEP/ SEP)
Course: Discrete Mathematics for ECE
Instructor: Engr. Jonalyn Mae E. Aranda
Law of Logical Equivalences
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 Law
pq  qp; pq  qp
Commutative Laws
(pq) r  p (qr); (pq)  r  p  (qr)
Associative Laws
Law of Logical Equivalences
p(qr)  (pq)(pr)
p(qr)  (pq)(pr)
Distribution Laws
(pq)(p  q)
(pq)(p  q)
De Morgan’s Laws
p  p  T
p  p  F
(pq)  (p  q)
Miscellaneous
Or Tautology
And Contradiction
Implication Equivalence
pq(pq)  (qp)
Biconditional Equivalence
Law of Logical Equivalences
Logical Equiavlences Involving Conditional Statements
(Implication)
Law of Logical Equivalences
Logical Equivalences Involving Biconditional Statements.
Law of Logical Equivalences
Note: De Morgan’s Laws extend to:
¬(p1 ∨ p2 ∨ ··· ∨ pn) ≡ (¬p1 ∧ ¬p2 ∧ ··· ∧ ¬pn)
and
¬(p1 ∧ p2 ∧ ··· ∧ pn) ≡ (¬p1 ∨ ¬p2 ∨ ··· ∨ ¬pn).
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.”
Answer
Use De Morgan’s laws to express the negations of “Miguel
has a cellphone and he has a laptop computer”
p: “Miguel has a cellphone”
q: “Miguel has a laptop computer.”
Then p ∧ q = “Miguel has a cellphone and he has a laptop
computer”. Negation: ¬(p ∧ q) is equivalent to ¬p ∨ ¬q.
Let:
¬p ∨ ¬q = “Miguel does not have a cellphone or he does
not have a laptop computer.”
Answer
“Heather will go to the concert or Steve will go to the
concert.”
r: “Heather will go to the concert”
s: “Steve will go to the concert.”
Then r ∨ s: “Heather will go to the concert or Steve will go
to the concert”.
Negation: ¬(r ∨ s) is equivalent to ¬r ∧ ¬s
Let
¬r ∧ ¬s = “Heather will not go to the concert and Steve
will not go to the concert.”
The Proof Process
Assumptions
Logical Steps
-Definitions
-Already-proved equivalences
-Statements (e.g., arithmetic
or algebraic)
Conclusion
(That which was to be proved)
Prove: (pq)  q  pq
(pq)  q
 q  (pq)
 (qp)  (q q)
 (qp)  T
 qp
 pq
Left-Hand Statement
Commutative
Distributive
Or Tautology
Identity
Commutative
Begin with exactly the left-hand side statement
End with exactly what is on the right
Justify EVERY step with a logical equivalence
Prove: (pq)  q  pq
(pq)  q
Left-Hand Statement
 q  (pq)
Commutative
(qp)  (q q) Distributive
Why did we need this step?
Our logical equivalence specified that  is distributive on the
right. This does not guarantee the equivalence works on the
left!
Ex.: Matrix multiplication is not always commutative
(Note that whether or not  is distributive on the left is not the
point here.)
Prove: p  q  q  p
pq
 p  q
 q  p
 (q)  p
 q  p
Contrapositive
Implication Equivalence
Commutative
Double Negation
Implication Equivalence
Prove: p  p  q is a tautology
Must show that the statement is true for any value of p,q.
ppq
 p  (p  q)
Implication Equivalence
 (p  p)  q
Associative
 (p  p)  q
Commutative
 Tq
Or Tautology
 qT
Commutative
T
Domination
This tautology is called the addition rule of
inference.
Why do I have to justify
everything?
• Note that your operation must have the
same order of operands as the rule you
quote unless you have already proven (and
cite the proof) that order is not important.
• 3+4 = 4+3
• 3/4  4/3
• A*B  B*A for everything (for example, matrix
multiplication)
Prove: (pq)  p is a tautology
(pq)  p
 (pq)  p
 (pq)  p
 (qp)  p
 q (p  p)
 q (p  p)
 q T
 T
Implication Equivalence
DeMorgan’s
Commutative
Associative
Commutative
Or Tautology
Domination
Prove or Disprove
p  q  p  q ???
To prove that something is not true it is
enough to provide one counter-example.
(Something that is true must be true in
every case.)
p q pq
pq
FT T
F
The statements are not logically equivalent
Prove:p  q p  q
p  q
 (pq)  (qp)
 (pq)  (qp)
 (pq)  (qp)
 (qp)  (pq)
 (qp)  (pq)
 (qp)  (pq)
 p  q
Biconditional Equivalence
Implication Equivalence (x2)
Double Negation
Commutative
Double Negation
Implication Equivalence (x2)
Biconditional Equivalence