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All the definitions I came across so far stated, that if a statement is true, then also its dual statement is true and this dual statement is obtained by changing + for ., 0 for 1 and vice versa.

However when I say 1+1, whose dual statement according to the above is 0.0, I get opposite results, that is:

1 + 1 = 1
0 . 0 = 0

How should I understand this duality principle ?

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6 Answers 6

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"$1 + 1 = 1$" is a statement (a boolean statement, in fact), and indeed, $1 + 1 = 1$ happens to be a true statement.

Likewise, the entire statement "$0 \cdot 0 = 0$" is a true statement, since $0 \cdot 0$ correctly evaluates to false: and this is exactly what "$0 \cdot 0 = 0$" asserts, so it is a correct (true) statement about the falsity of $0 \cdot 0$.

The duality principle ensures that "if we exchange every symbol by its dual in a formula, we get the dual result".

  • Everywhere we see 1, change to 0.
  • Everywhere we see 0, change to 1.
  • Similarly, + to $\cdot$, and $\cdot$ to +.

More examples:

(a) 0 . 1 = 0: is a true statement asserting that "false and true evaluates to false"

(b) 1 + 0 = 1: is the dual of (a): it is a true statement asserting that "true or false evaluates true."


(c) 1 . 1 = 1: it is a true statement asserting that "true and true evaluates to true".

(d) 0 + 0 = 0: (d) is the dual of (c): it is a true statement asserting, correctly, that "false or false evaluates to false".

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  • $\begingroup$ Ok, so I think I am slowly starting to get it, but I still have difficulties with the last sentence. The dual symbols are 1 and 0 and + and ., right ? And the dual result means, that if the original result was 1, the dual result will be 0. Is this correct ? $\endgroup$
    – jcxz
    Commented Jun 1, 2013 at 12:45
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    $\begingroup$ Yes, exactly. The "result" changes, but the truth value of the entire statement remains true. $\endgroup$
    – amWhy
    Commented Jun 1, 2013 at 12:48
  • $\begingroup$ ok, thanks very much, I think it is clear now $\endgroup$
    – jcxz
    Commented Jun 1, 2013 at 12:50
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    $\begingroup$ Good: I can understand the confusion! $\endgroup$
    – amWhy
    Commented Jun 1, 2013 at 12:51
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    $\begingroup$ Is duality principle applicable only when there is equality ? $\endgroup$ Commented Aug 14, 2015 at 12:26
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The statement is the full equation, including the = sign. 1+1 is neither true nor false: it takes the value 1, but it is not actually saying anything. Analogously, the expression "Tom has a cat" is neither true nor false (without specifying who Tom is) - it is an expression which could be true or false, depending on who we mean when we say "Tom".

On the other hand, the statement 1+1=0 is a false. Analogously, the statement "If Tom has a cat then Tom has no cats" is false, no matter who we mean when we say "Tom".

In this case, 1+1=1 is the true statement. Its dual is 0.0=0, which is also a true statement.

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  • $\begingroup$ Thanks for your explanation. I am no mathematician, so I need to have things explained in layman terms. So from your answer I understood that 1+1 is not a statement, so it makes no sense to compare its value to 0.0 and to talk about duality. On the other hand 1+1=1 and 0.0=0 are two true statements, so the duality principle is valid. Is this the correct understanding ? $\endgroup$
    – jcxz
    Commented Jun 1, 2013 at 12:33
  • $\begingroup$ 1 + 1 is indeed true, as a boolean statement, as the truth values are given [as true = 1], and + is defined as the boolean "or", so "true or true" has a truth value, and that happens to be true, just as 0.0 reads "false and false" which is false, so "false and false = false" is, as a whole, a true boolean statement. $\endgroup$
    – amWhy
    Commented Jun 1, 2013 at 12:40
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    $\begingroup$ An expression has a value, 1 or 0, but it is not a statement. As your question demonstrates, the duality principle only applies to statements, that is, things which contain an = sign. Perhaps it helps to think about 2+2 using normal addition. This clearly evaluates to 4, but we have no notion of whether it's true or not. However, 2+2=4 is clearly true. $\endgroup$ Commented Jun 1, 2013 at 15:36
  • $\begingroup$ Yes, indeed. I think I was just misunderstanding what you wrote: and the case here is a bit more complicated than addiction of integers, because here $1$ is identically true, and $0$ is identically false. $\endgroup$
    – amWhy
    Commented Jun 1, 2013 at 17:54
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The statement is not 1+1 but rather 1+1=1.

What the duality principle says is that "if you exchange every symbol by its dual in a formula you get the dual result".

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The difference between an expression and a statement is that statement is like an equation and expression is like variable.

"Duality of a statement is true" - What this means is that when you equate the dual of both the expressions in LHS and RHS of a statement(equation), the statement still makes sense.

Observe that in your example both 1+1 = 1 & 0.0 = 0 ,which are dual to each other, are true statements.

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The dual of $1.0$ which is $0+1$, will be obtained by interchanging $AND$ to $OR$ and $1$'s to $0$'s. The result of those function needn't be same.The results will be opposite to each other $1.0=0;0+1=1$, but they are the dual form of the single function $1.0$.
For example, the dual function of $XNOR$ is $XOR$.

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In my humble opinion, it is more helpful to describe the duality relationship in terms of variables. Example: Expression: x + x = x Dual of expression: x * x = x x could be 1 or 0. In all iterations, the expressions are correct. Good Luck!

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    $\begingroup$ Shouldn't the dual of $ x + x = x$ be $x' \cdot x'=x'$? $\endgroup$
    – Kaushik
    Commented Sep 20, 2019 at 19:55

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