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- From: Frédéric Blanqui <frederic.blanqui AT inria.fr>
- To: coq-club AT inria.fr
- Subject: Re: [Coq-Club] congruence of definitional equality
- Date: Thu, 30 Jan 2014 08:51:09 +0100
Le 30/01/2014 05:31, Abhishek Anand a
écrit :
I've seen the definitional equality of Coq being
presented I've never seen a proof that it is actually a
congruence. If so, why would such a least relation exist? Or, is it completely specified first and then proved to be a congruence? (e.g. in the style of http://www.cs.cornell.edu/Info/Projects/NuPrl/documents/Howe/howeEqualityinLazy_LICS98.ps) Or something else? Thanks, Hi. The definition is given on http://coq.inria.fr/distrib/current/refman/Reference-Manual006.html#sec171 : " Convertibility.Let us write E[Γ] ⊢ t ▷ u for the contextual closure of the relation t reduces to u in the environment E and context Γ with one of the previous reduction β, ι, δ or ζ. We say that two terms t1
and t2 are
βιδζη-convertible, or simply convertible, or equivalent,
in the environment E and
context Γ iff there exist terms u1
and u2 such
that
E[Γ] ⊢ t1 ▷ … ▷ u1 and
E[Γ] ⊢ t2 ▷ … ▷ u2 and either
u1 and u2 are identical, or
they are convertible up to
η-expansion, i.e. u1
is λ x:T. u′1 and u2 x is
recursively convertible to u′1,
or, symmetrically, u2
is λ
x:T. u′2 and u1 x is recursively convertible to
u′2. We
then write E[Γ] ⊢ t1 =βδιζη
t2."
Best regards, Frédéric. |
- [Coq-Club] congruence of definitional equality, Abhishek Anand, 01/30/2014
- Re: [Coq-Club] congruence of definitional equality, Jason Gross, 01/30/2014
- Re: [Coq-Club] congruence of definitional equality, Frédéric Blanqui, 01/30/2014
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