full equality check in coercion boundary #26
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Reference: rhi/quox#26
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e.g.
should work even tho 𝑖 is mentioned
also needs some kind of "known stuck" flag so it doesn't try it again and again if it hasn't changed
the rule in xtt (coercion regularity):
\frac{
Ψ, j, j' \mid Γ ⊢ A〈j/i〉 = A〈j'/i〉 ; \mathit{type}_k
}{
Ψ \mid Γ ⊢ \mathsf{coe} ; (i ⇒ A) ; p ; q ; s = (s ∷ A〈q/i〉) ⇒ A〈q/i〉
}
my current rule (shit):
\frac{
i ∉ \mathsf{fdv}(A)
}{
Ψ \mid Γ ⊢ \mathsf{coe} ; (i ⇒ A) ; p ; q ; s = (s ∷ A) ⇒ A
}
equivalent to the real rule that may or may not be easier:
\frac{
Ψ \mid Γ ⊢ A〈0/i〉 = A〈1/i〉 ; \mathit{type}_k
}{
Ψ \mid Γ ⊢ \mathsf{coe} ; (i ⇒ A) ; p ; q ; s = (s ∷ A〈1/i〉) ⇒ A〈1/i〉
}
test case: