My definition allows it to go backwards because for what im trying to do 
the order does not matter. So I evaluate the head of the sequence, and then 
recursively the tail.
I tested your example you gave me just now and it passed :D
Beatriz

A quarta-feira, 3 de fevereiro de 2021 à(s) 18:41:53 UTC, Robby Findler 
escreveu:

> Oh, I see -- you want (1 2 3) to be a subsequence of (1 4 2 4 3 4), for 
> example.
>
> But does your definition allow you to go "backwards"? Maybe you need a 
> helper judgment that captures "a subsequence that starts here"  and then 
> subsequence can try that one at each position?
>
> Robby
>
>
> On Wed, Feb 3, 2021 at 10:33 AM Beatriz Moreira <[email protected]> 
> wrote:
>
>> Yes, I had to make some adjustments to the judgement you sent me but this 
>> helped me a lot!
>> This was what I ended up using:
>>
>> (define-judgment-form L
>>   #:mode (subsequence I I)
>>   #:contract (subsequence (ts ...) (ts ...))
>>
>>   [----------------------------------------------
>>    (subsequence (ts_1 )
>>                 (ts_2 ... ts_1  ts_3 ...))]
>>   
>>   [(subsequence (ts_1 ...)
>>                 (ts_2 ... ts_3 ...))
>>    ----------------------------------------------
>>    (subsequence (ts_0 ts_1 ...)
>>                 (ts_2 ... ts_0 ts_3 ...))])
>>
>> Thank you so much! :D
>>
>> A sábado, 30 de janeiro de 2021 à(s) 20:08:17 UTC, Robby Findler escreveu:
>>
>>> Is this what you have in mind? 
>>>
>>> #lang racket
>>> (require redex/reduction-semantics)
>>>
>>> (define-language L
>>>   (ts ::= variable number))
>>>
>>> (define-judgment-form L
>>>   #:mode (subsequence I I)
>>>   #:contract (subsequence (ts ...) (ts ...))
>>>
>>>   [----------------------------------------------
>>>    (subsequence (ts_1 ...)
>>>                 (ts_2 ... ts_1 ... ts_3 ...))])
>>>
>>> (test-judgment-holds
>>>  (subsequence () (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (1) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (2) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (1 2) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (2 3) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (3 4) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (2 3 4) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (1 2 3) (1 2 3 4)))
>>>
>>> (test-judgment-holds
>>>  (subsequence (1 2 3 4) (1 2 3 4)))
>>>
>>> (test-equal
>>>  (judgment-holds (subsequence (5) (1 2 3 4))) #f)
>>>
>>> (test-equal
>>>  (judgment-holds (subsequence (3 2) (1 2 3 4))) #f)
>>>
>>> (test-equal
>>>  (judgment-holds (subsequence (4 1) (1 2 3 4))) #f)
>>>
>>> (test-results)
>>>
>>>
>>>
>>> On Sat, Jan 30, 2021 at 11:46 AM Beatriz Moreira <[email protected]> 
>>> wrote:
>>>
>>>> Hi !
>>>> I have a reduction relation where I have to match a pattern *ts* to 
>>>> two sequences, where the first one contains the other one. What I tried to 
>>>> do was something like this: 
>>>> 1st seq:  (*ts_all1 ... ts ts_all2 ...*)     2nd seq:   (*ts_x1 ... ts 
>>>> ts_x2 ...*),  where *ts_x* *⊆ **ts_all*.
>>>> But the problem is that with the pattern matching is that *ts_x1* 
>>>> doesn't match all elements of *ts_all1* .
>>>> I'm trying to use side conditions but i can't seem to get it right :c .
>>>> Any advice?
>>>>
>>>> Thank you , Beatriz :) 
>>>>
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>>>> .
>>>>
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>>
>

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