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jomboy
asked
in Digital Logic
Jul 14

356 views
5 votes

We use Xor Sign ->$\oplus$

We use Xnor sign->$\odot$

We have to confirm the associativity holds or not which mean we have to check ,

$\left ( \left ( a\oplus b \right )\odot c\right )$ =$\left ( a\oplus \left ( b\odot c \right ) \right )$ this holds not.

now we can confirm this by truth table.

Truth table for $\left ( \left ( a\oplus b \right )\odot c\right )$

$a$ | $b$ | $c$ | $a\oplus b$ | $\left ( \left ( a\oplus b \right )\odot c\right )$ |

0 | 0 | 0 | 0 | 1 |

0 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 1 | 0 |

0 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 1 | 0 |

1 | 0 | 1 | 1 | 1 |

1 | 1 | 0 | 0 | 1 |

1 | 1 | 1 | 0 | 0 |

Truth table for $\left ( a\oplus \left ( b\odot c \right ) \right )$

$a$ | $b$ | $c$ | $b\odot c$ | $\left ( a\oplus \left ( b\odot c \right ) \right )$ |

0 | 0 | 0 | 1 | 1 |

0 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 1 | 0 |

1 | 0 | 1 | 0 | 1 |

1 | 1 | 0 | 0 | 1 |

1 | 1 | 1 | 1 | 0 |

So it does not matter to which operator we associated b with $\odot$ or $\oplus$ the result of the boolean expression comes same for both the cases and it produce output 1 if the input is has even number of ones.