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- module Test where
- open import LF
- open import IID
- -- r←→gArgs-subst eliminates the identity proof stored in the rArgs. If this proof is
- -- by reflexivity r←→gArgs-subst is a definitional identity. This is the case
- -- when a = g→rArgs a'
- r←→gArgs-subst-identity :
- {I : Set}(γ : OPg I)(U : I -> Set)
- (C : (i : I) -> rArgs (ε γ) U i -> Set)
- (a' : Args γ U) ->
- let a = g→rArgs γ U a'
- i = index γ U a' in
- (h : C (index γ U (r→gArgs γ U i a))
- (g→rArgs γ U (r→gArgs γ U i a))
- ) -> r←→gArgs-subst γ U C i a h ≡ h
- r←→gArgs-subst-identity (ι i) U C _ h = refl-≡
- r←→gArgs-subst-identity (σ A γ) U C arg h = r←→gArgs-subst-identity (γ (π₀ arg)) U C' (π₁ arg) h
- where C' = \i c -> C i (π₀ arg , c)
- r←→gArgs-subst-identity (δ H i γ) U C arg h = r←→gArgs-subst-identity γ U C' (π₁ arg) h
- where C' = \i c -> C i (π₀ arg , c)
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