To prove that XOR (⊕) is the complement of XNOR (⊙), we will construct a truth table for both operations.
A | B | A ⊕ B (XOR) | A ⊙ B (XNOR) |
0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
From the table, we see that:
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When
𝐴⊕𝐵=0
then
𝐴⊙𝐵=1
When
𝐴⊕𝐵=1
then
𝐴⊙𝐵=0
Thus, XOR and XNOR are complements of each other. ✅
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