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Which of the following pairs of propositions are not logically equivalent?

  1. $((p \rightarrow r) \wedge (q \rightarrow r))$ and $((p \vee q) \rightarrow r)$
  2. $p \leftrightarrow q$ and $(\neg p \leftrightarrow \neg q)$
  3. $((p \wedge q) \vee (\neg p \wedge \neg q))$ and $p \leftrightarrow q$
  4. $((p \wedge q) \rightarrow r)$ and $((p \rightarrow r) \wedge (q \rightarrow r))$
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3 Answers

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A.  

$LHS. ((p→r) ∧ (q → r)) = ((p’+r) \ (q’+r)) = p’q’ + r $

$RHS. (p ∨ q) → r) = (p+q)’ + r = p’q’ +r = LHS$

 

B.

$LHS. \ p ↔ q = p ⊙ q$

$RHS. \  ¬p ↔¬q =  ¬p ⊙¬q = p⊙q = LHS $

 

C.

$LHS. \ (p ∧ q) ∨ (¬p ∧¬q) = pq + p’q’ = p ⊙ q = p ↔ q = RHS$

 

D.

$LHS. \ ((p ∧ q) → r) = pq → r = (pq)’+r = p’+q’+r$

$RHS. \ (p → r) ∧ (q → r) = (p’+r)(q’+r) = p’q’ + r ≠ LHS $

 

Correct Ans : Option D

 

 

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Logical Equivalences Involving Biconditional Statements.


p ↔ q ≡ (p → q) ∧ (q → p)
p ↔ q ≡ ¬p ↔¬q                        ( option 2)
p ↔ q ≡ (p ∧ q) ∨ (¬p ∧¬q)       (option 3)

 

Logical Equivalences Involving Conditional Statements

p → q ≡ ¬p ∨ q
p → q ≡ ¬q →¬p
¬(p → q) ≡ p ∧¬q
(p → q) ∧ (p → r) ≡ p → (q ∧ r)
(p → r) ∧ (q → r) ≡ (p ∨ q) → r  ( option 1)
(p → q) ∨ (p → r) ≡ p → (q ∨ r)
(p → r) ∨ (q → r) ≡ (p ∧ q) → r (in option 4  there is and instead of or )  hence correct ans is 4

(we can always  check via truth tables or by removing --> and <-> if above formulas are not learned properly)

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Ans :D

 ((p∧q)→r)((p∧q)→r) and ((p→r)∧(q→r))
Answer:

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