如何简化电路

Question

这个问题的答案是A，以防有人认为我考试作弊什么的。我真的不想明白如何解决这些问题。我猜想有某种方法可以将这些支架放在一起并获得简化的电路，但我不完全理解。也许是关于真值表的？

Answer 1

  abc' + a'bc' + a'b'c' + ab'c'        initial expression
= (a + a')bc' + a'b'c' + ab'c'         distributive law
= (T)bc' + a'b'c' + ab'c'              additive inverse
= bc' + a'b'c' + ab'c'                 multiplicative identity
= bc' + (a' + a)b'c'                   distributive law
= bc' + (T)b'c'                        additive inverse
= bc' + b'c'                           multiplicative identity
= (b + b')c'                           distributive law
= (T)c'                                additive inverse
= c'                                   multiplicative identity

表达式等价于c的逆，所以一个逻辑门就够了。