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[scott constable <sdconsta AT syr.edu>]

Fabian,

I also think you are correct that proving equality would require additional axioms. I don't have a problem with adding additional axioms; it would be useful to know which axiom(s) would be required to prove equality.

~Scott

On Wed, May 11, 2016 at 1:10 PM, Fabian Kunze <fkunze AT fakusb.de> wrote:

Hello Scott,

I'm pretty sure that, without additional axioms, you will just be able to prove "<->" instead of equality, e. g. "P ∘ Q ∘ R <-> (P ∘ Q) ∘ R."

Does this suffice?

Best,
Fabian

scott constable <sdconsta AT syr.edu> schrieb am Mi., 11. Mai 2016 19:07:

Correction: "I think it should be possible to prove associativity of my definition of composition"...

On Wed, May 11, 2016 at 1:06 PM, scott constable <sdconsta AT syr.edu> wrote:

Thanks for the quick response Frédéric, but I already have a large project which relies heavily on my definition of composition, and the CoLoR library uses a different definition of composition. Hence I would have to rework a lot of my project to fit the CoLoR definition. I think it should be possible to prove associativity of my definition of relation in Coq, and I would like to know how to do so.

~Scott

On Wed, May 11, 2016 at 1:00 PM, Frédéric Blanqui <frederic.blanqui AT inria.fr> wrote:

Hello. You will find useful developments on relations in CoLoR.Util.RelUtil.v (coq-color on opam). See http://color.inria.fr/doc/CoLoR.Util.Relation.RelUtil.html for the definitions and theorems without the proofs. Best regards, Frédéric.

Le 11/05/2016 18:51, scott constable a écrit :

Hi All,

I have the following definition of composition:

Inductive Compose : T → V → Prop :=
| comp_intro : ∀ x y z, R1 x y → R2 y z → Compose x z.

Notation "P ∘ R" := (Compose R P) (at level 55, right associativity).

And I am hopelessly lost as to how to prove this theorem:

Lemma compose_assoc : ∀ (R : U → V → Prop) (Q : V → T → Prop)
                          (P : T → W → Prop), P ∘ Q ∘ R = (P ∘ Q) ∘ R.

Any help would be extremely appreciated!

Thanks in advance,

~Scott Constable