It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.2K
- Applied Category Theory Course 343
- Applied Category Theory Exercises 149
- Applied Category Theory Discussion Groups 48
- Applied Category Theory Formula Examples 15
- Chat 475
- Azimuth Code Project 107
- News and Information 145
- Azimuth Blog 148
- Azimuth Forum 29
- Azimuth Project 190
- - Strategy 109
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 708
- - Latest Changes 700
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 20
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 38

Options

The string of inequalities in Eq. (2.15) is not quite a proof, because technically there is no such thing as \( v + w + u \), for example. Instead, there is \( (v+w)+u \) and \( v+(w+u) \), and so on.

- Formally prove, using only the rules of symmetric monoidal preorders ( Definition 2.2 ), that given the assertions in Eq. (2.13), the conclusion in Eq. (2.14) follows.
- Reflexivity and transitivity should show up in your proof. Make sure you are explicit about where they do.
- How can you look at the wiring diagram Eq. (2.6) and know that the commutative axiom (Definition 2.1 d.) does not need to be invoked?

**Eq 2.15** \( t + u \le v + w + u \le v + x + z \le y + z \)

**Eq 2.14** \( t + u \le y + z \)

**Eq 2.13** \(t \le v + w \) and \( w + u \le x + z \) and \( v + x \le y \)

## Comments

I tried to sketch a proof, but I don't know if it is correct. Let us start from the first assertion of (2.7), that is, \(t \le (v+w)\). We can write \((v+w) \le (v+w)+u\), that is `more rich.' By part (c) of Definition 2.1, we can write \((v+w)+u = v + (w+u)\). Using the second assertion in (2.7), we have: \(v + (w+u) \le v + (x+z)\). By (c), we get \(v + (x+z)=(v+x)+z\). Considering the third assertion, we can write \((v+x)+z\le y +z\). Let us go back to the first assertion: \(t \le (v+w)\); this implies \((t + u)\le (v+w)+u\). But \((v+w)+u\le y +z\), and thus, by transitivity, \((t + u)\le (y+z)\).

I have been thinking of other solutions. Wire \(t\) in diagram 2.4 branches into \(v+w\). I'm wondering if we might re-write \(t \le v + w\) as \(t+0 \le v+w\). I'm also wondering if we can consider this as the combination of \(t \le v\) and \(0 \le w\), `reversing' point (a) of Definition 2.1. In the same way, we can find \(w \le x\), \(u \le z\), as well as \(v \le y\) and \(x \le 0\). We have \(t \le v \le y\) and, by transitivity, \(t \le y\). Let us now just consider \(t \le y\) and \( u \le z\). By point (a), we get \(t+u \le y+z\), that is Eq. 2.8.

If we look at the wiring diagram, I guess that we do not need part (d) of Def. 2.1 because the wires do not twist.

`I tried to sketch a proof, but I don't know if it is correct. Let us start from the first assertion of (2.7), that is, \\(t \le (v+w)\\). We can write \\((v+w) \le (v+w)+u\\), that is `more rich.' By part (c) of Definition 2.1, we can write \\((v+w)+u = v + (w+u)\\). Using the second assertion in (2.7), we have: \\(v + (w+u) \le v + (x+z)\\). By (c), we get \\(v + (x+z)=(v+x)+z\\). Considering the third assertion, we can write \\((v+x)+z\le y +z\\). Let us go back to the first assertion: \\(t \le (v+w)\\); this implies \\((t + u)\le (v+w)+u\\). But \\((v+w)+u\le y +z\\), and thus, by transitivity, \\((t + u)\le (y+z)\\). I have been thinking of other solutions. Wire \\(t\\) in diagram 2.4 branches into \\(v+w\\). I'm wondering if we might re-write \\(t \le v + w\\) as \\(t+0 \le v+w\\). I'm also wondering if we can consider this as the combination of \\(t \le v\\) and \\(0 \le w\\), `reversing' point (a) of Definition 2.1. In the same way, we can find \\(w \le x\\), \\(u \le z\\), as well as \\(v \le y\\) and \\(x \le 0\\). We have \\(t \le v \le y\\) and, by transitivity, \\(t \le y\\). Let us now just consider \\(t \le y\\) and \\( u \le z\\). By point (a), we get \\(t+u \le y+z\\), that is Eq. 2.8. If we look at the wiring diagram, I guess that we do not need part (d) of Def. 2.1 because the wires do not twist.`

The approach suggested by Eq. 2.9 is that as a "cut line" moves from left to right as each inequality is "cut" a inequality is expressed. Put another way, the application of 2.1.a produces an implication. Throughout the proof \( + = \otimes \).

Given \( t + u \)

By 2.1.a \( x_1 = t , x_2 = u, y_1 = v + w , y_2 = u \) and 2.7.a \( t \le v + w \) giving \( t + u \le (v + w) + u \).

Applying 2.1.c we get \((v + w) + u = v + (w + u) \).

By 2.1.a \( x_1 = v, x_2 = w + u , y_1 = v, y_2 = x + z \) and 2.7.b \( w + u \le x + z \) giving \( v + (w + u) \le v + (x + z) \).

Applying 2.1.c we get \( v + (x + z) = (v + x) + z \).

By 2.1.a \( x_1 = v + x, x_2 = z, y_1 = y, y_2 = z \) and 2.7.c \( v + x \le y \) giving \( (v + x) + z \le y + z \).

`The approach suggested by Eq. 2.9 is that as a "cut line" moves from left to right as each inequality is "cut" a inequality is expressed. Put another way, the application of 2.1.a produces an implication. Throughout the proof \\( + = \otimes \\). Given \\( t + u \\) By 2.1.a \\( x_1 = t , x_2 = u, y_1 = v + w , y_2 = u \\) and 2.7.a \\( t \le v + w \\) giving \\( t + u \le (v + w) + u \\). Applying 2.1.c we get \\((v + w) + u = v + (w + u) \\). By 2.1.a \\( x_1 = v, x_2 = w + u , y_1 = v, y_2 = x + z \\) and 2.7.b \\( w + u \le x + z \\) giving \\( v + (w + u) \le v + (x + z) \\). Applying 2.1.c we get \\( v + (x + z) = (v + x) + z \\). By 2.1.a \\( x_1 = v + x, x_2 = z, y_1 = y, y_2 = z \\) and 2.7.c \\( v + x \le y \\) giving \\( (v + x) + z \le y + z \\).`

Thanks! This is very clear. I'm wondering if the first part of my sketch was correct, even if much longer.

`Thanks! This is very clear. I'm wondering if the first part of my sketch was correct, even if much longer.`

WIP

$$(v + w) + u \le (x + v) + z$$ $$(v + w) + u \le y + z$$ $$t + u \le y + z$$

`WIP $$(v + w) + u \le (x + v) + z$$ $$(v + w) + u \le y + z$$ $$t + u \le y + z$$`