What changes when moving from positive to negative logic for the same gate?

Every input and output bit is complemented

Only output bits are complemented

In the algebraic proof, Y_neg is obtained by:

Complementing both inputs and the output of the XOR

Complementing only the output of XOR

Complementing only the inputs of XOR

Asked 3/22
2070 Chaitra · 5 marks

Prove positive XOR equals negative XNOR

Short answer

In positive logic (H=1) the gate's output is XOR. In negative logic (H=0), H and L swap so the same physical gate yields 1 when inputs are equal, i.e., XNOR. Algebraically, Y_neg = (A' ⊕ B')' = A ⊙ B.

Prove that positive X-OR is equivalent to negative X-NOR.

In positive logic HIGH = 1 and LOW = 0; in negative logic HIGH = 0 and LOW = 1. Changing the convention complements every input and output bit of the same physical gate.

Take a gate that acts as XOR in positive logic. Its voltage table:

A	B	Y
L	L	L
L	H	H
H	L	H
H	H	L

Positive logic (H = 1, L = 0):

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	0

Y = 1 when inputs differ: XOR, $Y = A \oplus B$ .

Negative logic (H = 0, L = 1), same circuit:

A	B	Y
1	1	1
1	0	0
0	1	0
0	0	1

Rearranged in normal order (00, 01, 10, 11) the output is 1, 0, 0, 1, so Y = 1 when the inputs are equal: XNOR.

Algebraic proof: complement inputs and output of the positive-logic XOR:

\begin{aligned} Y_{neg} &= \overline{\bar A \oplus \bar B} \\ &= \overline{\bar A \bar{\bar B} + \bar{\bar A} \bar B} \\ &= \overline{\bar A B + A \bar B} \\ &= \overline{A \oplus B} = AB + \bar A \bar B = A \odot B \end{aligned}

Hence positive-logic XOR is equivalent to negative-logic XNOR.

✓ Answer checked by Claude Opus 5.5 (high effort)

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