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# Exercise 41 - Chapter 2

edited June 2018

Let $$\textbf{Bool} = ( \mathbb{B} , \le, true, \wedge)$$ and $$\textbf{Cost} = ([0, \infty], \ge, 0, +)$$ , and consider the following quasi-inverse functions $$d, u : [0, \infty] \rightarrow \mathbb{B}$$ defined as follows:

$$d(x) := \begin{cases} false & \text{if x > 0} \\ true & \text{if x = 0} \end{cases}$$ $$u(x) := \begin{cases} false & \text{if x = \infty } \\ true & \text{if x < \infty } \end{cases}$$ 1. Is d monotonic?
2. Does d satisfy conditions a. and b. of Definition 2.38?
3. Is d strict?
4. Is u monotonic?
5. Does u satisfy conditions a. and b. of Definition 2.38?
6. Is u strict?

• Options
1.

So we get $$u\dashv g\dashv d$$ as well.

Comment Source:So we get \$$u\dashv g\dashv d\$$ as well.
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2.
edited August 2018

$$f:P\to Q$$ is monotonic if(f) (see, e.g., Def. 1.50 in Seven Sketches): for every $$x,y\in P,\,\, x≤_Py \Rightarrow f(x)≤_Qf(y)$$.

Definition 2.38 copied from the link in the OP (with minor corrections):

Definition 2.34. Let $$\mathcal{P} = (P, \le_P , I_P , \otimes_P )$$ and $$\mathcal{Q} = (Q, \le_Q , I_Q , \otimes_Q )$$ be monoidal preorders. A monoidal monotone from $$\mathcal{P}$$ to $$\mathcal{Q}$$ is a monotone map $$f : (P, \le_P ) \rightarrow (Q, \le_Q )$$, satisfying two conditions:

a. $$I_Q ≤_Q f ( I_P )$$, and

b. $$f(p_1 ) \otimes_Q f(p_2 ) ≤_Q f (p_1 \otimes_P p_2 )$$

for all $$p_1 , p_2 \in P$$ .

strong monoidal monotone replace $$\le$$ with $$\cong$$

strict monoidal monotone replace $$\le$$ with $$=$$

1. $$\forall x,y\in [0,\infty]\,\mathrm{s.t.}\,x\geq y$$ one has that if $$d(y)=\mathrm{true}$$ then whatever the value of $$d(x)$$, $$d(x)\leq d(y)$$. If $$d(y)=\mathrm{false}$$, then $$y>0$$ so by transitivity $$x>0$$. $$x>0\Rightarrow d(x)=\mathrm{false}$$, making $$d(x)\leq d(y)$$ (in fact, they are equal). So, $$d:[0,\infty]\to\mathbb{B}$$ is monotonic.

2. a) $$I_\mathbb{B}=\mathrm{true}=d(0)=d(I_{[0,\infty]})$$. b) Let $$x,y\in[0,\infty]$$ be arbitrary. $$d(x)\otimes_\mathbb{B}d(y)=d(x)\wedge d(y)$$ So, if both $$d(x),d(y)=\mathrm{true},\,\,d(x)\otimes_\mathbb{B}d(y)=\mathrm{true}$$, otherwise it equals $$\mathrm{false}$$. $$d(x+y)=\mathrm{true}\Leftrightarrow x+y=0\Leftrightarrow x=y=0$$ in which case $$d(x)=d(y)=\mathrm{true}$$. Thus, $$d(x)\otimes_\mathbb{B}d(y)=d(x+y)=d(x\otimes_{[0,\infty]}y)$$. In conclusion, $$d:[0,\infty]\to\mathbb{B}$$ satisfies conditions a and b of Definition 2.38.

3. $$d:[0,\infty]\to\mathbb{B}$$ is strict since the inequalities in the definition are satisfied as equalities.

4. Let $$x\geq_{[0,\infty]}y$$. If $$u(y)=\mathrm{true}$$, then whatever $$u(x)$$ is $$u(x)\leq u(y)$$. If $$u(y)=\mathrm{false}$$, then $$y=\infty$$. By transitivity, $$x=\infty$$, so $$u(x)=\mathrm{false}=u(y)$$. Thus, $$u:[0,\infty]\to\mathbb{B}$$ is monotonic.

5. a) $$I_\mathbb{B}=\mathrm{true}=u(0)=u(I_{[0,\infty]})$$. b) Let $$x,y\in[0,\infty]$$ be arbitrary. $$u(x)\otimes_\mathbb{B}u(y)=u(x)\wedge u(y)$$. So, $$u(x),u(y)=\mathrm{true}$$ iff $$u(x)\otimes_\mathbb{B}u(y)=\mathrm{true}$$. $$u(x+y)=\mathrm{true}$$ iff $$x+y<\infty$$ iff $$x<\infty$$ and $$y<\infty$$ iff $$u(x)=u(y)=\mathrm{true}$$. So, $$u(x)\otimes_\mathbb{B}u(y)=u(x+y)$$, and $$u:[0,\infty]\to\mathbb{B}$$ satisfies conditions a and b.

6. Since conditions a and b are satisfied with equalities, $$u:[0,\infty]\to\mathbb{B}$$ is strict.

Comment Source:\$$f:P\to Q\$$ is monotonic if(f) (see, e.g., Def. 1.50 in Seven Sketches): for every \$$x,y\in P,\,\, x≤_Py \Rightarrow f(x)≤_Qf(y)\$$. Definition 2.38 copied from the link in the OP (with minor corrections): >**Definition 2.34**. Let \$$\mathcal{P} = (P, \le_P , I_P , \otimes_P ) \$$ and \$$\mathcal{Q} = (Q, \le_Q , I_Q , \otimes_Q ) \$$ be monoidal preorders. A monoidal monotone from \$$\mathcal{P} \$$ to \$$\mathcal{Q} \$$ is a monotone map \$$f : (P, \le_P ) \rightarrow (Q, \le_Q ) \$$, satisfying two conditions: >a. \$$I_Q ≤_Q f ( I_P ) \$$, and >b. \$$f(p_1 ) \otimes_Q f(p_2 ) ≤_Q f (p_1 \otimes_P p_2 ) \$$ >for all \$$p_1 , p_2 \in P \$$ . >**strong monoidal monotone** replace \$$\le\$$ with \$$\cong \$$ >**strict monoidal monotone** replace \$$\le\$$ with \$$= \$$ 1. \$$\forall x,y\in [0,\infty]\,\mathrm{s.t.}\,x\geq y\$$ one has that if \$$d(y)=\mathrm{true}\$$ then whatever the value of \$$d(x)\$$, \$$d(x)\leq d(y)\$$. If \$$d(y)=\mathrm{false}\$$, then \$$y>0\$$ so by transitivity \$$x>0\$$. \$$x>0\Rightarrow d(x)=\\mathrm{false}\$$, making \$$d(x)\leq d(y)\$$ (in fact, they are equal). So, \$$d:[0,\infty]\to\mathbb{B}\$$ is monotonic. 2. a) \$$I_\mathbb{B}=\mathrm{true}=d(0)=d(I_{[0,\infty]})\$$. b) Let \$$x,y\in[0,\infty]\$$ be arbitrary. \$$d(x)\otimes_\mathbb{B}d(y)=d(x)\wedge d(y)\$$ So, if both \$$d(x),d(y)=\mathrm{true},\,\,d(x)\otimes_\mathbb{B}d(y)=\mathrm{true}\$$, otherwise it equals \$$\mathrm{false}\$$. \$$d(x+y)=\mathrm{true}\Leftrightarrow x+y=0\Leftrightarrow x=y=0\$$ in which case \$$d(x)=d(y)=\mathrm{true}\$$. Thus, \$$d(x)\otimes_\mathbb{B}d(y)=d(x+y)=d(x\otimes_{[0,\infty]}y)\$$. In conclusion, \$$d:[0,\infty]\to\mathbb{B}\$$ satisfies conditions a and b of Definition 2.38. 3. \$$d:[0,\infty]\to\mathbb{B}\$$ is strict since the inequalities in the definition are satisfied as equalities. 4. Let \$$x\geq_{[0,\infty]}y\$$. If \$$u(y)=\mathrm{true}\$$, then whatever \$$u(x)\$$ is \$$u(x)\leq u(y)\$$. If \$$u(y)=\mathrm{false}\$$, then \$$y=\infty\$$. By transitivity, \$$x=\infty\$$, so \$$u(x)=\mathrm{false}=u(y)\$$. Thus, \$$u:[0,\infty]\to\mathbb{B}\$$ is monotonic. 5. a) \$$I_\mathbb{B}=\mathrm{true}=u(0)=u(I_{[0,\infty]})\$$. b) Let \$$x,y\in[0,\infty]\$$ be arbitrary. \$$u(x)\otimes_\mathbb{B}u(y)=u(x)\wedge u(y)\$$. So, \$$u(x),u(y)=\mathrm{true}\$$ iff \$$u(x)\otimes_\mathbb{B}u(y)=\mathrm{true}\$$. \$$u(x+y)=\mathrm{true}\$$ iff \$$x+y<\infty\$$ iff \$$x<\infty\$$ and \$$y<\infty\$$ iff \$$u(x)=u(y)=\mathrm{true}\$$. So, \$$u(x)\otimes_\mathbb{B}u(y)=u(x+y)\$$, and \$$u:[0,\infty]\to\mathbb{B}\$$ satisfies conditions a and b. 6. Since conditions a and b are satisfied with equalities, \$$u:[0,\infty]\to\mathbb{B}\$$ is strict.