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    • CommentRowNumber1.
    • CommentAuthorGuiGeek
    • CommentTimeSep 12th 2019

    In the page dependent sum type, the presented “term introduction rule” seems not really dependent, since the typing derivation of the second element is not aware of the first. I would expect t:Au:B[t/x](t,u):Σ(x:A).B\frac{\vdash t:A\quad \vdash u:B[t/x]}{\vdash (t,u):\Sigma(x:A).B} which corresponds to what is presented in the text on positive and negative versions.

    • CommentRowNumber2.
    • CommentAuthoratmacen
    • CommentTimeSep 12th 2019
    • (edited Sep 12th 2019)

    I guess they meant

    x:Xa:A(x)x:X(x,a): x:XA(x)\frac{x:X \vdash a:A(x)}{x:X \vdash (x,a) : \sum_{x':X}A(x')}

    which I think is basically the right idea. But that’s not the usual rule either, since it doesn’t allow you to construct pairs in the empty context, unless you have a primitive substitution rule.

    So the usual rule uses an arbitrary term of XX:

    t:Xa:A(t)(t,a): x:XA(x)\frac{t:X \qquad a:A(t)}{(t,a) : \sum_{x:X}A(x)}

    This is the same as what you wrote, but with different notation.

    • CommentRowNumber3.
    • CommentAuthorGuiGeek
    • CommentTimeSep 12th 2019

    But what they meant looks wrong to me, since it does not allow to introduce any pair which first element is not a variable, for instance to prove (0,Nil):Σ(n:Nat).Vectn(0,Nil):\Sigma(n:Nat).Vect\,n.

    • CommentRowNumber4.
    • CommentAuthoratmacen
    • CommentTimeSep 12th 2019

    Right, that’s why you’d need a substitution rule:

    Γa:AΓ,x:A,Γt:TΓ,Γ[a/x]t[a/x]:T[a/x]\frac{\Gamma \vdash a\,:\,A \qquad \Gamma,x:A,\Gamma' \vdash t\,:\,T}{\Gamma,\Gamma'[a/x] \vdash t[a/x]\,:\,T[a/x]}

    So combining this with the variables-only pairing rule (Γ\Gamma' \coloneqq \cdot, ata \coloneqq t, AXA \coloneqq X, t(x,a)t \coloneqq (x,a), T x:XA(x)T \coloneqq \sum_{x':X}A(x')) we get:

    Γt:XΓ,x:X(x,a): x:XA(x)Γ(t,a[t/x]): x:XA(x)\frac{\Gamma \vdash t\,:\,X \qquad \Gamma,x:X \vdash (x,a)\,:\,\sum_{x':X}A(x')}{\Gamma \vdash (t,a[t/x])\,:\,\sum_{x':X}A(x')}

    This is sufficiently general, since xx need not be free in aa. (And there’s actually no reason for xx to be free in aa when you just immediately substitute, since both tt and aa are provided together. What you really want is for xx to be free in A(x)A(x), of course.)

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeSep 12th 2019

    I think we should give the usual rule, as in #2. Anyone who wants to fix it, please do!

    • CommentRowNumber6.
    • CommentAuthoratmacen
    • CommentTimeSep 12th 2019

    Now I’m not sure anymore. The argument I gave justifies:

    t:Xx:Xa:A(x)(t,a[t/x]): x:XA(x)\frac{t:X \qquad x:X \vdash a\,:\,A(x)}{(t,a[t/x])\,:\,\sum_{x':X}A(x')}

    But having to prove A(x)A(x) generically, rather than just A(t)A(t) sounds harder, so maybe the rule is still too weak? Oh right, I guess it must be: what if XX is boolbool and A(x)A(x) is {true,false}\{true \mapsto \top,\;false \mapsto \bot\}?

    Yeah OK, it looks like requiring a variable was a red herring; the problem was that the premise isn’t depending on a specific term, like GuiGeek originally said.

    • CommentRowNumber7.
    • CommentAuthoratmacen
    • CommentTimeSep 12th 2019

    So the diagram next to the bad intro rule seems misleading: that xx arrow doesn’t have to be a context projection.

    • CommentRowNumber8.
    • CommentAuthoratmacen
    • CommentTimeSep 12th 2019
    Re #5: I edited the table include. Fixed?
    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTimeSep 12th 2019

    Yes, looks good. Thanks for catching this, GuiGeek, and for fixing it, Matt!

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