B A Y C P L E I T S C U PART II univalence n loop e w - - PowerPoint PPT Presentation

▶
b a y c p l e i t s c u
SMART_READER_LITE
LIVE PREVIEW

B A Y C P L E I T S C U PART II univalence n loop e w - - PowerPoint PPT Presentation

May 02, 2023 336 likes •571 views

B A Y C P L E I T S C U PART II univalence n loop e w ? p a t h s loop l o o p A By Jonah Kan - Own work, CC BY-SA 3.0, h ps://commons.wikimedia.org/w/index.php?curid=27584059 I know how to guarantee a combinatorial

slide-1
SLIDE 1

PART II

C U B I C A L E P Y T S

slide-2
SLIDE 2

l

  • p

A

n e w p a t h s

univalence loop

loop ?

slide-3
SLIDE 3

Daniel Marinus Kan

I know how to guarantee a combinatorial structure has enough paths

Thierry Coquand

By Andrej Bauer - hp://andrej.com/mathematicians/C/Coquand_Thierry.html, CC BY-SA 2.5 si, hps://commons.wikimedia.org/w/index.php?curid=15264235 By Jonah Kan - Own work, CC BY-SA 3.0, hps://commons.wikimedia.org/w/index.php?curid=27584059

My group knows how to bring that into the design

  • f type theory

*Guillaume Brunerie and Daniel R. Licata are also pioneers

slide-4
SLIDE 4

A

n e w p a t h s

univalence loop

n e w

  • p

e r a t

  • r

s f

  • r

e v e r y t y p e

  • 1. What are the types? (form)
  • 2. What are the constructors? (intro)
  • 3. How to consume an element? (elim)
  • 4. What if a constructor is consumed? (β)
  • 5. Uniqueness principle? (η)
  • 6. How to compose stu? (Kan operators)
slide-5
SLIDE 5

*Homogeneous Compositions

type

*not changing

type type cap face face

slide-6
SLIDE 6

should work with substitution [BCH]

slide-7
SLIDE 7

variant: diagonal faces and alternative filling directions

slide-8
SLIDE 8

Coercion

slide-9
SLIDE 9

variant 1: alternative coercion directions

slide-10
SLIDE 10

variant 2: freezing parts of the input

(used in cubical Agda)

slide-11
SLIDE 11

+

With these two operators every type has enough paths

slide-12
SLIDE 12

+

type

changing

They also give heterogeneous compositions

slide-13
SLIDE 13

Major Variants

[CCHM+CHM] [AFH+ABCFHL+CH] algebra on 𝕁 0, 1, ∧, ∨, ~ De Morgan 0, 1 coercion variant 2 variant 1 homogeneous composition standard variant ready-to-use proof assistants cubical Agda red

slide-14
SLIDE 14

i j i j i j k i j k

A : U M : A [ j=0 ↦ N1, i=1 ↦ N2, j=1 ↦ N3 ] 1 2 3

(faces are unordered in CCHM+CHM)

j=0 j=1 i=1

hcompk A [ j=0 ↦ N1, i=1 ↦ N2, j=1 ↦ N3 ] M : A

j=0, k:𝕁 ⊦ N1 : A [k=0 ↦ M] i=1, k:𝕁 ⊦ N2 : A [k=0 ↦ M, j=0 ↦ N1] j=1, k:𝕁 ⊦ N3 : A [k=0 ↦ M, i=1 ↦ N2]

homogeneous

slide-15
SLIDE 15

i j k

compk A [ … ] M : A[1/k]

k:𝕁 ⊦ A : U M : A[0/k]

slide-16
SLIDE 16

this represents (~i=1) = (i=0)

transpj A (~i) M : A[1/j]

i j

j:𝕁 ⊦ A : U M : A[0/j] i=0 ⊦ A ≡ A[0/j] : U

(type at i=0 cannot change)

in general, r:𝕁 to represent r=1

slide-17
SLIDE 17

φ ⊦ M : A

φ := false | true | (r = 0) | (r = 1) | φ1 ∧ φ2 | φ1 ∨ φ2 r := 0 | 1 | i | r1 ∧ r2 | r1 ∨ r2 | ~ r (De Morgan)

Constraints in Contexts

slide-18
SLIDE 18

φ := false | true | (r = 0) | (r = 1) | φ1 ∧ φ2 | φ1 ∨ φ2 r := 0 | 1 | i | r1 ∧ r2 | r1 ∨ r2 | ~ r (De Morgan)

Any φ is equivalent to r=1 for some r

e.g., (i=0)∨(i=1) = (~i=1)∨(i=1) = (~i∨i)=1

r ↦ r=1 preserves ∧, ∨, and ~ where ~r=1 means r=0

e.g., trapnsj A r M

slide-19
SLIDE 19

M : A [φ ↦ N] M : A φ ⊦ M ≡ N : A

and

Restricted by Partial Elements

slide-20
SLIDE 20
  • 2. What are the constructors? (intro)
  • 3. How to consume an element? (elim)
  • 4. What if a constructor is consumed? (β)
  • 5. Uniqueness principle? (η)
  • 6. hcomp and transp (and thus comp)

Gives us all the paths Definable for every type

slide-21
SLIDE 21

hcompi ⊤ [φ ↦ N] M ≡ M : ⊤ transpi ⊤ r M ≡ M : ⊤ M : ⊤ φ, i:𝕁 ⊦ N : ⊤ M : ⊤ r : 𝕁

slide-22
SLIDE 22

A

n e w p a t h s

univalence loop

hcomp transp

the unit the empty type functions pairs paths the circle universes

(many others)