Using truth table verify that ~ p ∨ q ≡ ~ p ∧ ~ q .

To verify that ~ p ∨ q ≡ ~ p ∧ ~ q p q p ∨ q ~ p ∨ q ~ p ~ q ~ p ∧ ~ q T T T F F F F T F T F F T F F T T F T F F F F F T T T T Hence, verified.

Find the symbolic form of the following switching circuit. Construct its switching table and interpret your result.

Given switching circuit is Let, p : The switch S 1 is closed. q : The switch S 2 is closed. ~ p : The switch S 1 ' is closed or the switch S 1 is open. ~ q : The switch S 2 ' is closed or the switch S 2 is open. ∴ The symbolic form of the given circuit is: p ∨ q ∧ ~ p ∧ ~ q So, the switching table is as follows: p q ~ p ~ q p ∨ q p ∨ q ∧ ~ p ∧ ~ q T T F F T F T F F T T F F T T F T F F F T T F F In the above truth table, all the entries in the last column are ' F ' , ∴ The given circuit represents a contradiction. ∴ Irrespective of whether switches S 1 and S 2 are open or closed, the given circuit will always be open (i.e., off). Hence, no current will flow in the circuit irrespective of the position of the switches.

Express the following circuit in symbolic form:

Let p : the switch S 1 is closed. q : the switch S 2 is closed. r : the switch S 3 is closed. ~ p : the switch S 1 ' is closed or the switch S 1 is open. ~ q : the switch S 2 ' is closed or the switch S 2 is open. ~ r : the switch S 3 ' is closed or the switch S 3 is open. l : the lamp L is on. The symbolic form of the given circuit is: p ∨ ( q ∧ r ) = l l is generally dropped and it can be written as: p ∨ ( q ∧ r ) Input-Output table p q r q ∧ r p ∨ ( q ∧ r ) 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0

Write the truth values of the following statement: 2 + 3 = 5 or 2 + 3 = 5 .

Given statement is 2 + 3 = 5 or 2 + 3 = 5 . The component statements are : Let p : 2 + 3 = 5 q : 2 + 3 = 5 Here, p is True T and q is False F . This is a Disjunction statement as the connecting word is 'or'. It is denoted as p ∨ q . So, p ∨ q is True T . Thus, the truth value of the given statement is True T .

Write the truth values of the following statement: 2 is a rational number and 2 is an irrational number.

Given statement is 2 is a rational number and 2 is an irrational number. The component statements are : Let p : 2 is a rational number. q : 2 is an irrational number. Here, p is True T and q is also True T . This is a Conjunction statement as the connecting word is 'and'. It is denoted as p ∧ q . So, p ∧ q is True T . Thus, the truth value of the given statement is True T .

Write the negation of the statement if diagonals of a parallelogram are perpendicular, then it is a rhombus.

The given statement is " If diagonals of a parallelogram are perpendicular, then it is a rhombus." Let p : diagonals of a parallelogram are perpendicular and q : it is a rhombus. So, the symbolic form is p → q The negation form of the symbolic statement will be ~ p → q = p ∧ ~ q i.e. "Diagonals of a parallelogram are perpendicular and it not a rhombus."

Write the negation of the statement a person is rich if and only if he is a software engineer.

The given statement is " A person is rich if and only if he is a software engineer." let p : A person is rich . q : He is a software engineer. Here ,the given statement is a biconditional statement with the symbolic form p ↔ q . So, the symbolic form of the negation of the statement will be , ~ p ↔ q = p ∧ ~ q ∨ q ∧ ~ p . ∴ The Negation of the given statement will be " A person is rich and he is not a software engineer or a person is software engineer and he is not rich."

Write the negation of the statement mangoes are delicious, but expensive.

The given statement is " Mangoes are delicious , but expensive." Let p : Mangoes are delicious. q : Mangoes are expensive. Here , the given statement is a conjunction statement. and the symbolic form of the given statement is p ∧ q . So, the symbolic form of the negation of the statement will be ~ p ∧ q = ~ p ∨ ~ q . ∴ The negation of the given statement will be " Mangoes are not delicious or they are not expensive."