Exercise A.1.6
Suppose you know that whenever X is true, then Y is true; that whenever Y is true, then Z is true; and whenever Z is true, then X is true. Is this enough to show that X, Y, Z are all logically equivalent? Explain.
As usual with these questions, let's build a truth table.
| X | Y | Z | X $\implies$ Y | Y $\implies$ Z | Z $\implies$ X | S |
|---|---|---|---|---|---|---|
| F | F | F | T | T | T | T |
| F | F | T | T | T | F | F |
| F | T | F | T | F | T | F |
| F | T | T | T | T | F | F |
| T | F | F | F | T | T | F |
| T | F | T | F | T | T | F |
| T | T | F | T | F | T | F |
| T | T | T | T | T | T | T |
The fiurth, fifth and sixth columns represent the conditions set out in the question, $X \implies Y$, $Y \implies Z$, and $Z \implies X$.
The final column S is the conjunction of all these conditions, and we can immediately recognise it as the logical equivalence of all three variables X, Y and Z.
So, yes, the conditions are sufficient to show X, Y and Z are logicallally equivalent. $\square$
It is interesting, and perhaps a pleasant surprise, to note that this kind of circular network of implications leads to logical equivalance. We have seen it in the simple case of 2 variables, where $X \implies Y$ and $Y \implies X$, is equivalent to $X \iff Y$.
Aside
It might be interesting to consider a network of implications with one or more closed loops, with appendages that are not part of any closed loop, to see which nodes (variables) are logically equivalent to the rest.
The above network shows that all the variables A-H are logically equivalent, but X is excluded from this logical equivalence.
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