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I have the logical formula $$ A \Leftrightarrow B \Leftrightarrow C $$ In order to make the truth table I'm not sure wheither I should interpret it as $A \Leftrightarrow B \Leftrightarrow C$ or $A \Leftrightarrow (B \Leftrightarrow C)$

Here is my first truth table:

$A$ $B$ $C$ $A \Leftrightarrow B \Leftrightarrow C$
1 1 1 1
0 1 1 0
1 0 1 0
0 0 1 0
1 1 0 0
0 1 0 0
1 0 0 0
0 0 0 1

I interpret $A \Leftrightarrow B \Leftrightarrow C$ true when all three are truth or all three are false

Here is my second truth table:

$A$ $B$ $C$ $A \Leftrightarrow (B \Leftrightarrow C)$
1 1 1 1
0 1 1 0
1 0 1 0
0 0 1 1
1 1 0 0
0 1 0 1
1 0 0 1
0 0 0 0

So is either one of theses forms the right one or am i all wrong ?

Nathaniel
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tixerauRIP
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2 Answers2

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The symbol $\Leftrightarrow$ is generaly defined as a binary operator. That means that $A\Leftrightarrow B\Leftrightarrow C$ is not defined without a priority between both $\Leftrightarrow$'s.

However, since $A\Leftrightarrow (B\Leftrightarrow C)$ and $(A\Leftrightarrow B)\Leftrightarrow C$ have the same truth tables, you can consider either of them. Another way to say it is that $\Leftrightarrow$ is associative.

Nathaniel
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There is no meaning in $A \Leftrightarrow B \Leftrightarrow C$ seen as a ternary operator. (By the way, how did you compute that truth table ??)

After placing parenthesis, the expression can be understood as "one variable is true and the other two equal, or that variable is false and the other two different". In all cases, an odd number of ones makes a true.

As this property is invariant to a permutation of the variables, associativity holds.

You can check by induction that this generalizes to more variables, and an expression like

$$ A \Leftrightarrow B \Leftrightarrow C \Leftrightarrow\cdots Z $$

is associative and corresponds to the parity of the number of true.