Hi Robin,
thank you for your comments. Indeed we are currently revising this
paper, hence your comments are especially welcome.
I've been reading "Towards Observational Type Theory". I'm new to type
theory, so I have not been able to understand big chunks of the paper;
hopefully as I read wider I'll understand more of it.
We should certainly strive to fix this...
However, one rule which jumped out at me is this, on page 9,
because it
seems silly (apologies if my email client mangles this even after I've
tried to fix it up - I've never tried pasting type rules from PDFs
into
it before):
Γ |- S≡T Γ |- Q:S=T Γ |- s:S
------------------------------
T
Γ |- s[Q>S ≡ s : T
It seems silly to include a definitional equality as a premise,
because
I _thought_ the whole point of definitional equality was that S≡T
means
that S and T are freely co-substitutable, and type-checking will
automatically substitute S for T or T for S _whenever_ required. (For
example, it is often noted that Coq will automatically reduce (i.e.
substitute) n+0 to n in a dependent type - but only if you write n+0
the "right" way round! I can't remember offhand which way round
works.)
You are right to say that this rule is unusual. In the implementation
it creates a dependency between defintional equaliy and reduction
which doesn't appear in conventional systems.
It can be justified by observing that in intensional Type Theory we
have tat
s [refl S> ≡ s
where refl S : S=S. However, due to proof-irrelevance any term Q : S
= S will be definitionally equal to refl S, hence if p:S=S then
s [p> ≡ s
So anyway, surely any derivation which needed that rule could also
use this rule instead:
Γ |- Q:S=S Γ |- s:S
------------------------------
S
Γ |- s[Q>S ≡ s : S
(which is trivial), substituting S for T before applying it, and
substituting Ts for some of the Ss afterwards.
You are right - your rule is equivalent and more concise. However, a
type checker will have to implement the first rule. To check whether
s [Q> reduces to s you have to verify whether the types in the type
of Q:S = T are definitionally equal, i.e. whether S ≡ T.
Where have I gone wrong?
Indeed nowhere.
There is some discussion of related issues in the Epigram developers
blog:
http://www.e-pig.org/epilogue/
is particular in the Epigram 2 design doc.
Cheers,
Thorsten
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