UNIT – 1
LOGIC
&
PROOFS
PROPOSITION
&
SYMBOLIC FORM
Symbolize the statement : “If Vani attends classes regularly
and if either she is attentive in the class or studies well then
she gets the top grade.”
ANS: Let P : Vani attends classes regularly.
Q : She is attentive in the class
R : She studies well
S : She gets the top grade.
The given statement can be written as P (Q R) → S
Write the following statement in symbolic form : If Avinash
is not in a good mood or he is not busy, then he will go to
New Delhi.
ANS : Let
P : Avinash is in a good mood
Q : He is busy
R : Go to New Delhi
Symbolic form : (¬𝑃 ∨ ¬Q) → R
State the truth value of “If tigers have wings then the earth
travels round the sun”.
ANS :
Let P : Tigers have wings have
truth value F
Q : The earth travels round the sun have truth value T
Therefore, P→Q has the truth value T.
TRUTH
TABLE
Construct the truth table 𝒑 ∧ 𝒒 → 𝒑 ∨ 𝐪 .
ANS :
p q 𝒑∧𝒒 𝒑∨𝒒 𝒑∧𝒒 → 𝒑∨𝒒
T T T T T
T F F T T
F T F T T
F F F F T
Construct the truth table for 𝑝 → 𝑞 ↔ ¬𝑝 → ¬𝑞 .
ANS :
Is ¬𝑝 ∧ 𝑝 ∨ 𝑞 →𝑞 a tautology.
ANS :
p q ¬𝒑 𝒑 ∨ 𝒒 (¬𝒑 (¬𝒑 ∧ 𝒑 ∨ 𝒒 )
∧ 𝒑∨𝒒 →𝒒
T T F T F T
T F F T F T
F T T T T T
F F T F F T
Prove that 𝑃 ∧ 𝑄 ∧ ¬(𝑃 ∨ 𝑄) is a contradiction.
ANS :
P Q 𝑷∧𝑸 𝑷∨𝑸 ¬(𝑷 ∨ 𝑸) 𝑷 ∧ 𝑸 ∧ ¬(𝑷
∨ 𝑸)
T T T T F F
T F F T F F
F T F T F F
F F F F T F
CONVERSE
CONTRAPOSITIVE
INVERSE
What are the contra positive, converse and inverse of the
conditional statement “If there is rain, then I buy an
umbrella”.
ANS : Let P : It rains today
Q : I buy an umbrella
Symbolic form : P→Q
Contrapositive : ¬Q → ¬𝑃
Converse : Q→P
Inverse : ¬P → ¬Q
What are the contra positive , converse and inverse of the
conditional statement “ If you work hard then you will be
rewarded”.
ANS : Let P : You work hard
Q : You will be rewarded
Symbolic form : P→Q
Contrapositive : ¬Q → ¬𝑃
Converse : Q→P
Inverse : ¬P → ¬Q
Write the converse and contra-positive of the conditional
statement: If you obey the traffic rules, then you will not
be fined.
ANS: Let P : You obey the traffic rules
Q : You will not be fined.
The given statement : P → Q.
Converse : Q → P.
Contra-positive : 7Q → 7P.
LOGICAL
IMPLICATION
Prove that 𝑝 → 𝑞 ∧ 𝑞 → 𝑟 ⟹ 𝑝 → 𝑟
ANS: Antecedent : Assume that 𝑝 → 𝑞 ∧ 𝑞 → 𝑟 is true
Consequent : To prove 𝑝 → 𝑟 is true.
𝑝 → 𝑞 ∧ 𝑞 → 𝑟 is true ⟹ 𝑝 → 𝑞 𝑖𝑠 𝑡𝑟𝑢𝑒 & 𝑞 → 𝑟 is true
𝑝 → 𝑞 is true ⟹ 𝑝 is true & 𝑞 is true --------(1)
𝑝 is false & 𝑞 is true --------(2)
𝑝 is false & 𝑞 is false --------(3)
𝑞 → 𝑟 is true ⟹ 𝑞 is true & 𝑟 is true --------(4)
𝑞 is false & 𝑟 is true --------(5)
𝑞 is false & 𝑟 is false --------(6)
From (1) &(4) ⟹ 𝑝 → 𝑟 is true
From (2) &(5) ⟹ 𝑝 → 𝑟 is true
From (3) &(6) ⟹ 𝑝 → 𝑟 is true.
Hence 𝑝 → 𝑞 ∧ 𝑞 → 𝑟 ⟹ 𝑝 → 𝑟
Prove that 𝑃 → 𝑄 ∧ 𝑅 → 𝑄 ⇒ 𝑃 ∨ 𝑅 → 𝑄.
ANS:
𝑃→𝑄 ∧ 𝑅→𝑄
⟺ (¬𝑃 ∨ 𝑄) ∧ (¬𝑅 ∨ 𝑄)
⟺ (¬𝑃 ∧ ¬𝑅) ∨ 𝑄
⟺ ¬ (𝑃 ∨ 𝑅) ∨ 𝑄
⟺ 𝑃∨𝑅 →𝑄
LOGICAL
EQUIVALENCE
Show that 𝑃 → 𝑄 → 𝑅 ⟺ 𝑃 ∧ 𝑄 → 𝑅 ⟺ 𝑃 → ¬𝑄 ∨ 𝑅 .
ANS:
𝑃→ 𝑄→𝑅 ⟺ ¬𝑃 ∨ 𝑄 → 𝑅
⟺ ¬𝑃 ∨ ¬𝑄 ∨ 𝑅
⟺ (¬𝑃 ∨ ¬𝑄) ∨ 𝑅
⟺ ¬(𝑃 ∧ 𝑄) ∨ 𝑅
⟺ 𝑃∧𝑄 →𝑅
𝑃 → 𝑄 → 𝑅 ⟺ ¬𝑃 ∨ 𝑄 → 𝑅
⟺ ¬𝑃 ∨ ¬𝑄 ∨ 𝑅
⟺ 𝑃 → ¬𝑄 ∨ 𝑅
Prove without using truth table
¬ 𝑝 ⟷ 𝑞 ≡ 𝑝 ∨ 𝑞 ∧ ¬ 𝑝 ∨ 𝑞 ≡ 𝑝 ∧ ¬𝑞 ∨ ¬𝑝 ∧ 𝑞 .
ANS: ¬ 𝑝⟷𝑞 ⟺¬ 𝑝→𝑞 ∧ 𝑞→𝑝
⟺ ¬ ¬𝑝 ∨ 𝑞 ∧ ¬𝑞 ∨ 𝑝
⟺ ¬ (¬𝑝 ∨ 𝑞 ∧ ¬𝑞) ∨ ¬𝑝 ∨ 𝑞 ∧ 𝑝
⟺ ¬ ¬𝑝 ∧ ¬𝑞 ∨ (q ∧ ¬𝑞) ∨ ¬𝑝 ∧ 𝑝) ∨ (𝑞 ∧ 𝑝
⟺ ¬ ¬𝑝 ∧ ¬𝑞 ∨ F ∨ F ∨ 𝑞 ∧ 𝑝
⟺ ¬ ¬𝑝 ∧ ¬𝑞 ∨ 𝑞 ∧ 𝑝
⟺ 𝑝 ∨ 𝑞 ∧ (¬𝑞 ∨ 𝑞)
⟺ 𝑝∨𝑞 ∧¬ 𝑝∨𝑞 PTO
¬ 𝑝⟷𝑞 ⟺ ¬ 𝑝→𝑞 ∧ 𝑞→𝑝
⟺ ¬ ¬𝑝 ∨ 𝑞 ∧ ¬𝑞 ∨ 𝑝
⟺ 𝑝 ∧ ¬𝑞 ∨ ¬𝑝 ∧ 𝑞
Prove that 𝑝⋁𝑞 ⋀¬ ¬𝑝⋀ ¬𝑞⋁¬𝑟 ⋁ ¬𝑝⋀¬𝑞 ⋁ ¬𝑝⋀¬𝑟 is a
tautology.
ANS: 𝑝⋁𝑞 ⋀¬ ¬𝑝⋀ ¬𝑞⋁¬𝑟 ⋁ ¬𝑝⋀¬𝑞 ⋁ ¬𝑝⋀¬𝑟
⟺ 𝑝 ∨ 𝑞 ∧ ¬ ¬𝑝⋀¬ 𝑞 ∧ 𝑟 ∨ ¬ 𝑝 ∨ 𝑞 ∨ ¬(𝑝 ∨ 𝑟)
⟺ 𝑝∨𝑞 ∧ 𝑝∨ 𝑞∧𝑟 ∨ ¬[ 𝑝 ∨ 𝑞 ∧ (𝑝 ∨ 𝑟)]
⟺ 𝑝 ∨ 𝑞 ∧ 𝑝 ∨ 𝑞 ∧ (𝑝 ∨ 𝑟) ∨ ¬[𝑝 ∨ 𝑞 ∧ 𝑟 ]
⟺ 𝑝∨𝑞 ∧ 𝑝∨𝑟 ∨ ¬[𝑝 ∨ 𝑞 ∧ 𝑟 ]
⟺ [𝑝 ∨ (𝑞 ∧ 𝑟)] ∨ ¬[𝑝 ∨ 𝑞 ∧ 𝑟 ]
⟺ T
PCNF
&
PDNF
Obtain PCNF & PDNF of 𝑝 → 𝑞 ∧ 𝑟 ∧ ¬𝑝 → ¬𝑞 ∧ ¬𝑟 .
ANS: 𝑝→ 𝑞∧𝑟 ∧ ¬𝑝 → ¬𝑞 ∧ ¬𝑟
⟺ [¬𝑝 ∨ 𝑞 ∧ 𝑟 ] ∧ [𝑝 ∨ ¬𝑞 ∧ ¬𝑟 ]
⟺ ¬𝑝 ∧ 𝑝 ∨ ¬𝑝 ∧ ¬𝑞 ∧ ¬𝑟 ∨ 𝑞 ∧ 𝑟 ∧ 𝑝 ∨ (𝑞 ∧ 𝑟 ∧ ¬𝑞 ∧ ¬𝑟)
⟺ F ∨ ¬𝑝 ∧ ¬𝑞 ∧ ¬𝑟 ∨ (𝑝 ∧ 𝑞 ∧ 𝑟) ∨ 𝐹
PDNF of S ⟺ ¬𝑝 ∧ ¬𝑞 ∧ ¬𝑟 ∨ (𝑝 ∧ 𝑞 ∧ 𝑟)
PDNF of ¬S ⟺ 𝑝 ∧ 𝑞 ∧ ¬𝑟 ∨ 𝑝 ∧ ¬𝑞 ∧ 𝑟 ∨ 𝑝 ∧ ¬𝑞 ∧ ¬𝑟 ∨ ¬𝑝 ∧ 𝑞 ∧ 𝑟
∨ ¬𝑝 ∧ 𝑞 ∧ ¬𝑟 ∨ (¬𝑝 ∧ ¬𝑞 ∧ 𝑟)
PCNF OF S ⟺ ¬[ PDNF of ¬S]
PCNF OF S ⟺ ¬𝑝 ∨ ¬𝑞 ∨ 𝑟 ∧ ¬𝑝 ∨ 𝑞 ∨ ¬𝑟 ∧ ¬𝑝 ∨ 𝑞 ∨ 𝑟
∧ 𝑝 ∨ ¬𝑞 ∨ ¬𝑟 ∧ 𝑝 ∨ ¬𝑞 ∨ 𝑟 ∧ 𝑝 ∨ 𝑞 ∨ ¬𝑟)
Obtain PDNF of 𝑃 ∧ 𝑄 ∨ 𝑅 → ¬𝑃 & hence find its PCNF.
ANS: 𝑃 ∧ 𝑄 ∨ 𝑅 → ¬𝑃 ⟺ ¬[ 𝑃 ∧ 𝑄 ∨ 𝑅] ∨ ¬𝑃 ⟺ [ ¬𝑃 ∨ ¬𝑄 ∧ ¬𝑅] ∨ ¬𝑃
⟺ ¬𝑃 ∨ ¬𝑄 ∨ ¬𝑃 ∧ ¬𝑅 ∨ ¬𝑃
⟺ ¬𝑃 ∨ ¬𝑄 ∧ ¬𝑅 ∨ ¬𝑃 ⟺ (¬𝑃 ∨ ¬𝑄) ∨ 𝐹 ∧ (¬𝑅 ∨ ¬𝑃) ∨ 𝐹
⟺ ¬𝑃 ∨ ¬𝑄 ∨ (𝑅 ∧ ¬𝑅) ∧ ¬𝑅 ∨ ¬𝑃 ∨ (𝑄 ∧ ¬𝑄)
⟺ ¬𝑃 ∨ ¬𝑄 ∨ 𝑅 ∧ (¬𝑃 ∨ ¬𝑄 ∨ ¬𝑅) ∧ ¬𝑅 ∨ ¬𝑃 ∨ 𝑄 ∧ (¬𝑅 ∨ ¬𝑃 ∨ ¬𝑄)
PCNF of S ⟺ ¬𝑃 ∨ ¬𝑄 ∨ 𝑅 ∧ (¬𝑃 ∨ ¬𝑄 ∨ ¬𝑅) ∧ ¬𝑃 ∨ 𝑄 ∨ ¬𝑅
PCNF of ¬S ⟺ 𝑃 ∨ 𝑄 ∨ 𝑅 ∧ ¬𝑃 ∨ 𝑄 ∨ 𝑅 ∧ 𝑃 ∨ ¬𝑄 ∨ 𝑅
∧ 𝑃 ∨ 𝑄 ∨ ¬𝑅 ∧ 𝑃 ∨ ¬𝑄 ∨ ¬𝑅
PDNF of S ⟺ ¬( PCNF of ¬S )
PDNF of S ⟺ 𝑃 ∧ ¬𝑄 ∧ ¬𝑅 ∨ ¬𝑃 ∧ 𝑄 ∧ 𝑅 ∨ ¬𝑃 ∧ 𝑄 ∧ ¬𝑅
∨ ¬𝑃 ∧ ¬𝑄 ∧ 𝑅 ∨ ¬𝑃 ∧ ¬𝑄 ∧ ¬𝑅 .
Show that ¬𝑃 → 𝑅 ∧ 𝑄 ⟷ 𝑃
= 𝑃⋁𝑄⋁𝑅 ⋀ 𝑃⋁¬𝑄⋁𝑅 ⋀ 𝑃⋁¬𝑄⋁¬𝑅 ⋀ ¬𝑃⋁𝑄⋁𝑅 ⋀ ¬𝑃⋁𝑄⋁¬𝑅 .
ANS:
¬𝑃 → 𝑅 ∧ 𝑄 ⟷ 𝑃 ⟺ ¬𝑃 → 𝑅 ∧ 𝑄 → 𝑃 ∧ (𝑃 → 𝑄)
⟺ 𝑃 ∨𝑅 ∧ ¬𝑄 ∨ 𝑃 ∧ (¬𝑃 ∨𝑄)
⟺ 𝑃 ∨𝑅 ∨𝐹 ∧ ¬𝑄 ∨𝑃 ∨𝐹 ∧ [(¬𝑃 ∨𝑄) ∨ 𝐹]
⟺ 𝑃 ∨𝑅 ∨(𝑄 ∧ ¬𝑄) ∧ ¬𝑄 ∨ 𝑃 ∨(𝑅 ∧ ¬𝑅) ∧ [(¬𝑃 ∨𝑄) ∨ (𝑅 ∧ ¬𝑅)]
⟺ 𝑃 ∨𝑅 ∨ 𝑄 ∧ 𝑃 ∨𝑅 ∨¬𝑄 ∧ ¬𝑄 ∨𝑃 ∨𝑅 ∧ ¬𝑄 ∨𝑃 ∨¬𝑅
∧ (¬𝑃 ∨𝑄 ∨𝑅) ∧ (¬𝑃 ∨𝑄 ∨¬𝑅)
⟺ 𝑃 ∨𝑄 ∨ 𝑅 ∧ 𝑃 ∨ ¬𝑄 ∨𝑅 ∧ 𝑃 ∨¬𝑄 ∨¬𝑅 ∧ (¬𝑃 ∨𝑄 ∨𝑅) ∧ (¬𝑃 ∨𝑄 ∨¬𝑅)
INFERENCE
THEORY
Show that 𝑝 → 𝑞 ∧ 𝑟 → 𝑠 , 𝑞 → 𝑡 ∧ 𝑠 → 𝑢 , ¬ 𝑡 ∧ 𝑢 and 𝑝 → 𝑟 ⟹ ¬𝑝.
ANS:
Premises Rule Reason
1. 𝑝→𝑞 ∧ 𝑟→𝑠 P Given Premises
2. 𝑝→𝑞 T (1) simplification
3. 𝑟→𝑠 T (1) simplification
4. 𝑞→𝑡 ∧ 𝑠→𝑢 P Given Premises
5. 𝑞→𝑡 T (4) simplification
6. 𝑠→𝑢 T (4) simplification
7. 𝑝→𝑡 T (2), (5) Hypothetical
8. 𝑟→𝑢 T (3),(6) Hypothetical
9. 𝑝→𝑟 P Given Premises
10. 𝑝→𝑢 T (9), (8) Hypothetical
11. ¬𝑢 → ¬𝑝 T (10) contrapositive
12. ¬𝑡 → ¬𝑝 T (7) contrapositive
13. (¬𝑡 ∨ ¬𝑢) → ¬𝑝 T (12), (11)
𝑝 → 𝑞 ; 𝑟 → 𝑞 ⇒ (𝑝 ∨ 𝑟) → 𝑞
14. ¬(𝑡 ∧ 𝑢) → ¬𝑝 T (13) (¬(𝑝 ∧ 𝑞) ⇒ ¬𝑝 ∨ ¬𝑞
Demorgans
15. ¬(𝑡 ∧ 𝑢) P Given Premises
16. ¬𝑝 T (15), (16) Modus ponens
Show that the conclusion R follows from P → Q , Q → R , P ∨ R by using
indirect method.
ANS:
Premises Rule Reason
1. ¬𝑅 P Added Premises
2. 𝑄→𝑅 P Added Premises
3. ¬𝑄 T (2),(1) Modus Tollens
4. 𝑃→𝑄 P Given Premises
5. ¬𝑃 T (4), (3) Modus Tollens
6. 𝑃∨𝑅 P Given Premises
7. ¬𝑃 → 𝑅 T (6) Implication
8. 𝑅 T (5),(7) Modus ponens
9. 𝑹 ∧ ¬𝑹 = F T (1),(8)
Prove that 𝐴 → ¬𝐷 is a conclusion from the premises 𝐴 → 𝐵 ∨ 𝐶, 𝐵 → ¬𝐴 and D→
¬𝐶 by using conditional proof.
ANS :
Premises Rule Reason
1. 𝐴 P Added Premises
2. 𝐴→𝐵∨𝐶 P Given Premises
3. 𝐵∨𝐶 T (1),(2) Modus ponens
4. ¬𝐵 → 𝐶 T (3) implication
5. 𝐵 → ¬𝐴 P Given Premises
6. 𝐴 → ¬𝐵 T (5) contrapositive
7. 𝐴→𝐶 T (6), (4) Hypothetical
8. D→ ¬𝐶 P Given Premises
9. ¬𝐷⋁𝐶 T (8) implication
10. ¬𝐷 T (9) Addition 𝑝⋁𝑞 ⟺ 𝑝
11. 𝐴 → ¬𝐷 CP (1), (10)
Determine the validity of the following argument: My father praises me only if I can
be proud of myself. Either I do well in sports or I can’t be proud of myself. If I study
hard, then I can’t do well in sports. Therefore if father praises me, then I do not study
well.
ANS: Let 𝒑 : My father praises me 𝒒 : I can be proud of myself
𝒓 : I do well in sports 𝒔 : I study well
Premises : 𝒑 → 𝒒 , 𝒓 ∨ ¬𝒒 , 𝒔 → ¬𝒓 Conclusion : 𝒑 → ¬𝒒
Premises Rule Reason
1. 𝑝→𝑞 P Given Premises
2. 𝑝 P Added Premises
3. 𝑞 T (1),(2) Modus ponens
4. 𝑟 ∨ ¬𝑞 P Given Premises
5. 𝑟 T (3),(4) Disjunctive
6. 𝑠 → ¬𝑟 P Given Premises
7. 𝑟 → ¬𝑠 T (6) Contrapostive
8. ¬𝑠 P (5),(7) Modus ponens
9. 𝒑 → ¬𝒒 CP (2), (8)
Show that the following premises are inconsistent. “If Ram misses many classes through illness
then he fails high school. If Ram fails high school then he is uneducated. If Ram reads a lot of
books then he is not uneducated. Ram misses many classes through illness and reads a lot of
books.
ANS: Let 𝑝 : Ram misses many classes 𝑞 : Ram fails high school
𝑟 : Ram reads lot of books 𝑠 : Ram is uneducated
Premises : 𝒑 → 𝒒 , 𝒒 → 𝒔 , 𝒓 → ¬𝒔 , 𝒑 ∧ 𝒓
Premises Rule Reason
1. 𝑝→𝑞 P Given Premises
2. 𝑞→𝑠 P Given Premises
3. 𝑝→𝑠 T (1), (2) Hypothetical
4. 𝑟 → ¬𝑠 P Given Premises
5. 𝑠 → ¬𝑟 T (4) contrapositive
6. 𝑝 → ¬𝑟 T (3), (5) Hypothetical
7. ¬𝑝 ∨ ¬𝑟 = ¬( 𝑝 ∧ 𝑟) T (6) implication
8. 𝑝∧𝑟 P Given Premises
𝑝∧𝑟 ∧¬ 𝑝∧𝑟 = T (7),(8)
F
QUANTIFIERS
Symbolize: Every book with a blue cover is a
mathematics book.
ANS : Let P(x) : x is a book
B(x) : x has a blue cover
M(x) : x is a mathematics book
for all x, if x is a book and x has a blue cover then x
is a mathematics book
Symbolic form :
⇒ ∀x P(x) ∧ B(x) ⟶ M(x) .
Write the symbolic form & negate the following statement
(i)Everyone who is healthy can do all kinds of work.
(ii) Some people are not admired by everyone.
(iii) Everyone should help his neighbors or his neighbors will not help him.
ANS:
(i) Let 𝐻 𝑥 ∶ 𝑥 is healthy
𝑊 𝑥 ∶ 𝑥 can do all kinds of work
Symbolic form 𝑥 𝐻 𝑥 →𝑊 𝑥
(i) Let 𝐴 𝑥 ∶ 𝑥 is admired
Symbolic form ∃𝑥 ¬𝐴 𝑥
(i) Let 𝐻 𝑥 ∶ 𝑥 help neighbor
𝑃 𝑥 ∶ 𝑥 is person
Symbolic form 𝑥 [(𝑃 𝑥 → 𝐻 𝑥 ) ∨ (𝐻 𝑥 → ¬𝑃 𝑥 )]
Show that ∃𝑥𝑃 𝑥 ∧ ∃𝑥𝑄 𝑥 and ∃𝑥 𝑃 𝑥 ∧ 𝑄 𝑥 are not logically
equivalent.
ANS:
Let us assume that 𝑃 𝑥 ∶ 𝑥 is a positive integer
𝑄 𝑥 ∶ 𝑥 is a negative integer
By truth table ∃𝑥𝑃 𝑥 ∧ ∃𝑥𝑄 𝑥 is true.
[ ∵ ∃𝑥𝑃 𝑥 𝑖𝑠 𝑡𝑟𝑢𝑒 & ∃𝑥𝑄 𝑥 𝑖𝑠 𝑡𝑟𝑢𝑒 ]
But the truth value of 𝑃 𝑥 ∧ 𝑄 𝑥 is false
Hence ∃𝑥[ 𝑃 𝑥 ∧ 𝑄 𝑥 ] is false
∴ ∃𝑥𝑃 𝑥 ∧ ∃𝑥𝑄 𝑥 and ∃𝑥 𝑃 𝑥 ∧ 𝑄 𝑥 are not logically equivalent
Using CP (or) otherwise prove that ∀𝑥 𝑃 𝑥 → 𝑄 𝑥 , ∀𝑥 [𝑅 𝑥 →
¬𝑄 𝑥 ] ⟹ ∀𝑥 𝑅 𝑥 → ¬𝑃 𝑥 .
ANS:
S.No Premises Rule Reason
1. ∀𝑥 𝑃 𝑥 → 𝑄 𝑥 P Given premises
2. 𝑃 𝑦 →𝑄 𝑦 US ∀𝑥 𝑃(𝑥) ⇒ 𝑃(𝑦)
3. ∀𝑥 𝑅 𝑥 → ¬𝑄 𝑥 P Given premises
4. 𝑅 𝑦 → ¬𝑄 𝑦 US ∀𝑥 𝑃(𝑥) ⇒ 𝑃(𝑦)
5. 𝑄 𝑦 →¬𝑅 𝑦 P (4) contrapositive
6. 𝑃 𝑦 →¬𝑅 𝑦 T 2 , 5 hypothetical
7. 𝑅 𝑦 →¬𝑃 𝑦 T (6) Contrapostive
8. ∀𝑥 𝑅 𝑥 → ¬𝑃 𝑥 UG (8)
Show that the premises “One student in this class knows how to write programs in JAVA”
and “Everyone who knows how to write programs in JAVA can get high paying job”
imply the conclusion “Someone in this class can get a high paying job”.
ANS: Let A(x) : x is in the class B(x) : x knows JAVA programming
H(x) : x can get a high paying job.
Premises : ∃𝑥 𝐴 𝑥 ∧ 𝐵 𝑥 , 𝑥 𝐵 𝑥 →𝐻 𝑥 Conclusion : ∃𝑥 𝐴 𝑥 ∧ 𝐻 𝑥
S.No Premises Rule Reason
1. ∃𝑥 𝐴 𝑥 ∧ 𝐵 𝑥 P Given premises
2. 𝐴 𝑦 ∧𝐵 𝑦 ES ∃𝑥 𝑃(𝑥) ⇒ 𝑃(𝑦)
3. 𝐴 𝑦 T P∧Q ⇒ P
4. 𝐵 𝑦 T P∧Q ⇒ Q
5. 𝑥 𝐵 𝑥 →𝐻 𝑥 P Given premises
6. 𝐵 𝑦 →𝐻 𝑦 US ∀𝑥 𝑃(𝑥) ⇒ 𝑃(𝑦)
7. 𝐻 𝑦 T (4), (6) Modus ponens
8. 𝐴 𝑦 ∧𝐻 𝑦 T (3), (4)
9. ∃𝑥 𝐴 𝑥 ∧ 𝐻 𝑥 EG (8)
Verify the validity of the following argument. Lions are dangerous animals. There are Lions.
Therefore, there are dangerous animals.
ANS:
Let (𝒙) : 𝒙 is a lion Q(𝒙) : 𝒙 is a dangerous animal
Premises : (∀𝒙)[P 𝒙 → 𝑸(𝒙)] , ∃𝒙 𝑷 𝒙
Conclusion : ∃𝒙 𝑸 𝒙
S.No Premises Rule Reason
1. (∀𝑥)[P 𝑥 → 𝑄(𝑥)] P Given premises
2. P 𝑦 → 𝑄(𝑦) US ∀𝑥 𝑃(𝑥) ⇒ 𝑃(𝑦)
3. ∃𝑥 𝑃 𝑥 P Given premises
4. P 𝑦 ES ∃𝑥 𝑃 𝑥 ⇒ 𝑃(𝑦)
5. 𝑄(𝑦) T (4), (2) Modus Ponens
6. ∃𝑥 𝑄 𝑥 EG 𝑃(𝑦) ⇒ ∃𝑥 𝑃 𝑥
