Question

Consider the following Boolean expression of a function F :
𝐹(𝑃, 𝑄) = (𝑃̅ + 𝑄) ⊕(𝑃̅𝑄)
Which of the following expressions is/are equivalent to F ?

• A. 𝑃⊕𝑄 ̅̅̅̅̅̅̅̅
• B. 𝑃⊕𝑄
• C. 𝑃̅ ⊕𝑄
• D. 𝑃̅ ⊕𝑄̅

Hint:

Build the truth table of F for all four combinations of P and Q, then compare it row by row against each option; F turns out to be the XNOR of P and Q.

Answer 1

The Correct Option is A, C

The given Boolean function is \( F(P, Q) = (\overline{P} + Q) \oplus (\overline{P}Q) \). We need to test each option against the truth table of F.

Step 1: Construct the truth table for F.

For each combination of P and Q, compute \(\overline{P} + Q\) and \(\overline{P}Q\), then XOR them.

• P=0, Q=0: \(\overline{P}=1\), so \(\overline{P}+Q=1\), \(\overline{P}Q=0\). \(F = 1 \oplus 0 = 1\)
• P=0, Q=1: \(\overline{P}=1\), so \(\overline{P}+Q=1\), \(\overline{P}Q=1\). \(F = 1 \oplus 1 = 0\)
• P=1, Q=0: \(\overline{P}=0\), so \(\overline{P}+Q=0\), \(\overline{P}Q=0\). \(F = 0 \oplus 0 = 0\)
• P=1, Q=1: \(\overline{P}=0\), so \(\overline{P}+Q=1\), \(\overline{P}Q=0\). \(F = 1 \oplus 0 = 1\)

So F equals 1, 0, 0, 1 for (P,Q) = (0,0), (0,1), (1,0), (1,1) respectively. This is exactly the truth table of the XNOR function, that is \(F = \overline{P \oplus Q}\).

Step 2: Check option (A) \(\overline{P \oplus Q}\).

\(P \oplus Q\) is 0,1,1,0 for the same four rows, so its complement is 1,0,0,1, which matches F exactly. Option A is equivalent to F.

Step 3: Check option (B) \(P \oplus Q\).

This gives 0,1,1,0, which is the exact complement of F, not F itself. Option B is not equivalent.

Step 4: Check option (C) \(\overline{P} \oplus Q\).

Evaluate at each row: (0,0): \(1 \oplus 0 = 1\); (0,1): \(1 \oplus 1 = 0\); (1,0): \(0 \oplus 0 = 0\); (1,1): \(0 \oplus 1 = 1\). This gives 1,0,0,1, matching F exactly. Option C is equivalent to F.

Step 5: Check option (D) \(\overline{P} \oplus \overline{Q}\).

Since complementing both inputs of an XOR does not change its value, \(\overline{P} \oplus \overline{Q} = P \oplus Q\), which is 0,1,1,0, not matching F.

Therefore, the expressions equivalent to F are options A and C.

\[ \boxed{\text{Options A and C}} \]