DeMorgan's law and C++

Question

For each of the following write the equivalent C++ expressions, without any unary negation operators (!). (!= is still permitted)

Use DeMorgan\'s law

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Answer 1

Check this out:

!(x!=5 && x!=7)                 -->    x==5 || x==7

!(x<5 || x>=7)                  -->    x>=5 && x<7

!( !(a>3 && b>4) && (c != 5))   -->    (a>3 && b>4) || c==5

So, just #2 from your solutions is correct.