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# Exercise 31 - Chapter 2

edited June 2018

Consider the preorder $$(P, \le )$$ with Hasse diagram $$\Big[ \text{ no } \le \text{ maybe } \le \text{ yes } \Big]$$ .

We propose a monoidal structure with "yes" as the monoidal unit and "min" as the monoidal product.

1) Make sense of "min" by filling in the multiplication table with elements of P.

2) Check the axioms of Definition 2.2 hold for $$\mathbf{NMY} := (P, \le, \text{yes}, \text{min} )$$, given your definition of "min".

$$\begin{array}{c|ccc} \text{min} & no & maybe & yes \\ \hline no & ? & ? & ? \\ maybe & ? & ? & ? \\ yes & ? & ? & ? \end{array}$$ Previous Next

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1.

The multiplication table is

$$\begin{array}{c|ccc} \text{min} & no & maybe & yes \\ \hline no & no & no & no \\ maybe & no & maybe & maybe \\ yes & no & maybe & yes \end{array}$$ The "min" operation is associative and commutative, so properties (c) and (d) of Definition 2.1 hold.

For any element $$x \in P, x \otimes yes = yes \otimes x = \min{(x, yes)} = x$$, so the unit laws (b) also hold.

Suppose $$x_1\leq y_1, x_2\leq y_2$$. Since $$P$$ is a chain, we must have $$x_1\leq x_2$$ and/or $$x_2\leq x_1$$; consider the former case (the other case is dealt with similarly).

We need to show that $$x_1\leq y_1 \otimes y_2 =\min{(y_1, y_2)}$$. This is true by assumption if $$y_1 =\min{(y_1, y_2)}$$, so consider the case $$y_2 =\min{(y_1, y_2)}$$.

We then have either $$x_1 \leq y_2$$ or $$x_2 \leq y_2 < x_1 \leq y_1$$. If the latter were true, then transitivity would imply $$x_2 < x_1$$ contrary to the assumption that $$x_1\leq x_2$$. So we must have $$x_1 \leq y_2$$. This establishes property (a) of Definition 2.1 and thus concludes the proof.

Comment Source:The multiplication table is $$\begin{array}{c|ccc} \text{min} & no & maybe & yes \\\\ \hline no & no & no & no \\\\ maybe & no & maybe & maybe \\\\ yes & no & maybe & yes \end{array}$$ The "min" operation is associative and commutative, so properties (c) and (d) of Definition 2.1 hold. For any element \$$x \in P, x \otimes yes = yes \otimes x = \min{(x, yes)} = x\$$, so the unit laws (b) also hold. Suppose \$$x_1\leq y_1, x_2\leq y_2\$$. Since \$$P\$$ is a chain, we must have \$$x_1\leq x_2\$$ and/or \$$x_2\leq x_1\$$; consider the former case (the other case is dealt with similarly). We need to show that \$$x_1\leq y_1 \otimes y_2 =\min{(y_1, y_2)}\$$. This is true by assumption if \$$y_1 =\min{(y_1, y_2)}\$$, so consider the case \$$y_2 =\min{(y_1, y_2)}\$$. We then have either \$$x_1 \leq y_2\$$ or \$$x_2 \leq y_2 < x_1 \leq y_1\$$. If the latter were true, then transitivity would imply \$$x_2 < x_1 \$$ contrary to the assumption that \$$x_1\leq x_2\$$. So we must have \$$x_1 \leq y_2\$$. This establishes property (a) of Definition 2.1 and thus concludes the proof.
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2.

It seems to be suggested that there's some logical connective correspondence for this min (as in Example 2.10 and Exercise 2.11). Is that what question 2 means by "definition of min"? :-?

Comment Source:It seems to be suggested that there's some logical connective correspondence for this _min_ (as in Example 2.10 and Exercise 2.11). Is that what question 2 means by "definition of min"? :-?