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## Here's the Solution to this Question

1)

i) You can become successful and happy unless you lead a lazy life.

define P = You can become successful and happy and Q = lead a lazy lazy life

Rewriting the sentence we have

If you can become successful and happy then you have to avoid being lazy in life.

inverse

if you cannot become successful and happy then you do not have to avoid being lazy in life.

converse

if you lead a lazy life then you can become successful and happy.

contrapositive

if you not lead a lazy life then you cannot become successful and happy.

ii) We enter the password or we click on the forget password button is necessary for

us to login into the system and create a user profile.

Define P = We enter the password or we click on the forget password button

Define Q = log into the system and create a user profile

Rewriting the sentence, we have

If we enter the password or we click on the forget password button then we can login into the system and create user profile.

inverse

If we do not enter the password or we do not click on the forget password button then we cannot login into the system and create user profile

converse

if we can log in to the system and create a user profile then we enter the password or we click on forget password button

contrapositive

if we cannot log into the system and create user profile then we do not enter the password or we do not click on the forget password button

2) Define if you study= a, get good grades = b, did assignments = c

Thus the three specifications can be reduced to

i) a$\implies$ b

ii) c$\implies$b

iii) $\lnot$ c

when both a, b and c are false, then all the three statements are true. So the specification is consistent.

3. Use any of the two proof methods to prove the following- [5]

∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) ≡ a

assuming that a = True (T) and b= False (F), then

(∼a ∧ b) =( ∼T∧ F ) = F

(∼a ∧ ∼b) =( ∼T ∧ ∼F ) = F

Thus (∼a ∧ b) ∨ (∼a ∧ ∼b) = (F∨ F) = F

Hence ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) = ∼F = T

(a ∧ b) = (T ∧ F) = F

Thus ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) =( T∨ F) = T

Since a = T, this means the left side(∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) is equivalent to the right side(a). Hence we conclude that ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) ≡ a which concludes the proof.

Again, assuming a = false(F) and b= True(T), then

(∼a ∧ b) =( ∼F∧ T ) = T

(∼a ∧ ∼b) =( ∼F ∧ ∼T ) = F

Thus (∼a ∧ b) ∨ (∼a ∧ ∼b) = (T∨ F) = T

Hence ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) = ∼T = F

(a ∧ b) = (F ∧ T) = F

Thus ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) =( F∨ F) = F

Since a = False, this means the left side(∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) is equivalent to the right side(a). Hence we conclude that ∼((∼a ∧ b) ∨ (∼a ∧ ∼b)) ∨ (a ∧ b) ≡ a which concludes the proof.

By the two proofs above where we have used different logical values for a and b, we conclude the proof as shown above in both proofs.