A K N Correction: We will redo the empty type the rules were - PowerPoint PPT Presentation
A K N Correction: We will redo the empty type the rules were wrong; the video was re-uploaded hcomp i ( x:A B) [ N] M N(x) M(x) N(x) x.hcomp i B [ N(x)] M(x) B X fill the square with the same x along i A i transp i
A K N
Correction: We will redo the empty type the rules were wrong; the video was re-uploaded
hcomp i ( Π x:A B) [ φ ↦ N] M N(x) M(x) N(x) λ x.hcomp i B [ φ ↦ N(x)] M(x) B X fill the square with the same x along i A i
transp i ( Π x:A B) φ M λ x.transp i B[filler i (x)/x] φ M(filler 0 (x)) M(x') M(x) filler i (x) : ≡ transp-fill ~i A φ x filler x back and forth x' i
hcomp i ( Σ x:A B) [ φ ↦ N] M snd(N) ⟨ hcomp i A [ φ ↦ fst(N)] fst(M) , s n comp i B[filler i /x] [ φ ↦ snd(N)] snd(M) ⟩ d snd(N) ( M ) filler i : ≡ hfill i A [ φ ↦ fst(N)] fst(M) fst(N) filler needs comp i because fst(M) fst(N) B(filler/x) depends on i i
transp i ( Σ x:A B) φ M ⟨ transp i A φ fst(M) , snd(M) transp i B[filler i /x] φ snd(M) ⟩ filler filler i : ≡ transp-fill i A φ fst(M) fst(M) i
M OPERATORS r r e e l l l l i i f f M DERIVED A A i i transp-fill ~i A φ M transp-fill i A φ M N N filler filler M M N N A A i i i hfill i A [ φ ↦ N] M comp i A [ φ ↦ N] M
the unit natural numbers functions disjoint sums pairs the empty type paths the circle universes
Q Q N P Q P M Q Path j.A (M; N) A hcomp i (Path j.A (M; N)) [ φ ↦ Q] P ≡ λ j.hcomp i A [ φ ↦ Q@j, j=0 ↦ M, j=1 ↦ N] P@j
transp i (Path j.A (M; N)) φ P ≡ λ j.comp i A [ φ ↦ P@j, j=0 ↦ M, j=1 ↦ N] P@j N P P M Path j.A (M; N) A
the unit natural numbers functions disjoint sums pairs the empty type paths the circle universes negative types positive types
transp i ℕ φ M M M always works
N hcomp i ℕ [ φ ↦ N] M M If φ = true, this should reduce to N[1/i] N Otherwise, what should we do? Freely generated inductive types now have irreducible hcomps
C ℕ elim(hcomp) = comp(elim) E(hcomp i ℕ [ φ ↦ O] P) ≡ comp i C[filler i /x] [ φ ↦ E(O)] E(P) E(O) : ≡ elim ℕ [x.C](M; x.y.N; O) filler i : ≡ hfill i ℕ [ φ ↦ O] P
1. Inductive types have formal hcomps 2. Elim commutes with formal hcomps e.g., ⊥ , ℕ , 2, A+B, the circle, etc. transp can always be reduced* *except for indexed inductive families
inr inl A B A+B transp (A+B) φ inl(M) ≡ inl(transp A φ M) transp (A+B) φ inr(M) ≡ inr(transp B φ M)
Optional* hcomp can commute with constructors ℕ suc(hcomp i ℕ [ φ ↦ N] M) ≡ hcomp i ℕ [ φ ↦ suc(N)] suc(M) A+B inl(hcomp i A [ φ ↦ N] M) ≡ hcomp i (A+B) [ φ ↦ inl(N)] inl(M) inr(hcomp i B [ φ ↦ N] M) ≡ hcomp i (A+B) [ φ ↦ inr(N)] inr(M) *Cubical Agda has these rules
the unit natural numbers functions disjoint sums pairs the empty type paths the circle universes negative types positive types
Recommend
More recommend
Explore More Topics
Stay informed with curated content and fresh updates.