'Difference between if (j%3 != 0 && j%4 != 0) and (!((j%3 == 0) && (j%4 == 0))), and why (!((j%3 == 0) || (j%4 == 0))) works if || means OR
I don't understand why the following syntax are not the same
if (j%3 != 0 && j%4 != 0)
if (!((j%3 == 0) && (j%4 == 0)))
Yet, the followings are.
if (j%3 != 0 && j%4 != 0) /
if (!((j%3 == 0) || (j%4 == 0)))
Why is that
!(A && B) != !A && !B
and
!(A || B) == !A && !B / !(A && B) == !A || !B
Solution 1:[1]
This is the mathematic boolean logic.
A and B <=> not (not A or not B)
not A and not B <=> not (A or B)
not (A and B) <=> not A or not B
For proving these equations, use value tables.
0 & 0 = 0 !(1 | 1) = !1 = 0
0 & 1 = 0 !(1 | 0) = !1 = 0
1 & 0 = 0 !(0 | 1) = !1 = 0
1 & 1 = 1 !(0 | 0) = !0 = 1
Solution 2:[2]
Say I have a red car and grey house, then "my car is blue and my house is grey" is false (!(A && B)). Yet "my car isn't blue and my house isn't grey" is also false (!(!A && !B)). So !(A && B) and !A && !B clearly aren't equivalent.
The actual negation of "my car is blue and my house is grey" is "my car isn't blue or my house isn't grey" (!A || !B).
We can also clearly see this is the case by plugging all possible values into the equations.
| A | B | !(A && B) | !A || !B | !(A || B) | !A && !B |
|---|---|---|---|---|---|
| 0 | 0 | !(0 && 0) = !0 = 1 | !0 || !0 = 1 || 1 = 1 | !(0 || 0) = !0 = 1 | !0 && !0 = 1 && 1 = 1 |
| 0 | 1 | !(0 && 1) = !0 = 1 | !0 || !1 = 1 || 0 = 1 | !(0 || 1) = !1 = 0 | !0 && !1 = 1 && 0 = 0 |
| 1 | 0 | !(1 && 0) = !0 = 1 | !1 || !0 = 0 || 1 = 1 | !(1 || 0) = !1 = 0 | !1 && !0 = 0 && 1 = 0 |
| 1 | 1 | !(1 && 1) = !1 = 0 | !1 || !1 = 0 || 0 = 0 | !(1 || 1) = !1 = 0 | !1 && !1 = 0 && 0 = 0 |
Two equivalences that are identified by the above table:
!(A && B) = !A || !B
!(A || B) = !A && !B
These are known as De Morgan's laws.
Sources
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Source: Stack Overflow
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| Solution 1 | |
| Solution 2 |