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- From: Conor McBride <conor AT strictlypositive.org>
- To: coq-club AT pauillac.inria.fr
- Subject: Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?
- Date: Thu, 14 Feb 2008 13:51:52 +0000
- List-archive: <http://pauillac.inria.fr/pipermail/coq-club/>
Hi
I sent a message to Gang Chen when I meant to send it
to the list, but it basically says what Bruno says...
On 14 Feb 2008, at 11:54, Bruno Barras wrote:
In fact, Gang Chen's property (GC) is equivalent to propositional
extensionality
PE : forall A B: Prop,(A<->B)->A=B.
Obviously PE -> GC.
GC->PE is also provable:
Axiom equiv : forall (F : Prop->Prop) (A B : Prop), (A<->B)->((F A) <->(F
B)).
Goal forall A B, (A<->B) -> A=B.
intros.
elim (equiv (fun P => P=B) _ _ H); intros.
auto.
Qed.
PE is (believed) consistent with Coq since there is (but this is not
formally established yet) a boolean model of Coq (i.e. where Prop is
interpreted by booleans), and PE is valid in this model.
(And this shows that the godelisation above cannot be effective without
axiom.)
However PE is (believed) not provable,
...and adds a more syntactic account of non-provability
of PE, which I'll repeat and expand.
If PE held in Coq, we could use it to prove that, for
some trivial closed proposition T, we have
T = T -> T
in the empty context. But reflexvity cannot construct
such a proof.
In Epigram 2's core theory, propositional equality
(defined type by type, extensional for functions) on
propositions is defined to be logical equivalence. Of
course, we then have to ensure that there are no
possible observations on propositions which allow finer
distinctions. It's rather nice, because if we have an
equivalence relation R, then
R x y <-> R x = R y
But I digress.
All the best
Conor
- [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?, gang chen
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Vincent Aravantinos
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Robin Green
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Vincent Aravantinos
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Gyesik Lee
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?, Andreas Abel
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Bruno Barras
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?, Conor McBride
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?, gang chen
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Gyesik Lee
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Vincent Aravantinos
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Robin Green
- Re: [Coq-Club] is "(A<->B)->((F A)<->(F B))" provable ?,
Vincent Aravantinos
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