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Exercise 41 - Chapter 2

edited June 2 in Exercises

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:

$$ d(x) := \begin{cases} false & \text{if $x > 0$} \\ true & \text{if $x = 0$} \end{cases} $$ $$ u(x) := \begin{cases} false & \text{if $x = \infty $} \\ true & \text{if $x < \infty $} \end{cases} $$ Figure

  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?

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Comments

  • 1.

    So we get \(u\dashv g\dashv d\) as well.

    Comment Source:So we get \\(u\dashv g\dashv d\\) as well.
  • 2.
    edited August 1

    \(f:P\to Q\) 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.
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