I think I get it ...
> REWRITE_RULE [FUN_EQ_THM] (ASSUME ``R = R'``);
<<HOL message: inventing new type variable names: 'a>>
val it =
[.] |- R = R':
thm
> REWRITE_RULE [FUN_EQ_THM] (ASSUME ``(R:'a -> 'a -> bool) = R'``);
val it =
[.] |- ∀x x'. R x x' ⇔ R' x x':
thm
> Il giorno 24 mar 2017, alle ore 21:50, Thomas Tuerk <tho...@tuerk-brechen.de>
> ha scritto:
>
> Hi Chun,
>
> use functional extensionality. There are many ways to do it, one is using
> the theorem boolTheory.FUN_EQ_THM.
>
> Best
>
> Thomas
> On 24.03.2017 21:42, Chun Tian (binghe) wrote:
>> Hi again,
>>
>> If I have a theorem saying two (2-ary) relations are the same:
>>
>> |- R = R’
>>
>> Then I can easily prove the following theorem using REWRITE_TAC:
>>
>> |- !x y. R x y = R’ x y
>>
>> But if I had the second one first, how to prove the previous one?
>>
>> Regards,
>>
>> Chun Tian
>>
>>
>>
>>
>>
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