Sometimes Not Question

hotranchsauce

Are the following two equivalent?

(E1) --(s)-- (-E2)
(E2) --(s)-- (-E1)

canihazJD

edit... it looks like you are saying this?

A ←s→ /B
B ←s→ /A

In which case, no they are not.

hotranchsauce

@canihazJD said:
edit... it looks like you are saying this?

A ←s→ /B
B ←s→ /A

In which case, no they are not.

Thanks. Let me ask one more question if I can.

If A is not always B, then A<--s-->/B, correct?

So then can we or can't we conclude that /A<--s-->B?

I thought that because "some" goes both ways, that you could put the "/" on either side and have it be equally true. But I obviously have not convinced myself because I'm here talking about it.

Thanks again for shining any light onto this, I really appreciate it.

canihazJD

@businesskarafa said:

@canihazJD said:
edit... it looks like you are saying this?

A ←s→ /B
B ←s→ /A

In which case, no they are not.

Thanks. Let me ask one more question if I can.

If A is not always B, then A<--s-->/B, correct?

So then can we or can't we conclude that /A<--s-->B?

I thought that because "some" goes both ways, that you could put the "/" on either side and have it be equally true. But I obviously have not convinced myself because I'm here talking about it.

Thanks again for shining any light onto this, I really appreciate it.

/A ←s→ B might very well be true, but you can't infer it just from A ←s→ /B.

/A and A are just two sets that don't intersect. They also could just be called △and ☐, △ including everything that is not a ☐.

hotranchsauce

@canihazJD said:

@businesskarafa said:

@canihazJD said:
edit... it looks like you are saying this?

A ←s→ /B
B ←s→ /A

In which case, no they are not.

Thanks. Let me ask one more question if I can.

If A is not always B, then A<--s-->/B, correct?

So then can we or can't we conclude that /A<--s-->B?

I thought that because "some" goes both ways, that you could put the "/" on either side and have it be equally true. But I obviously have not convinced myself because I'm here talking about it.

Thanks again for shining any light onto this, I really appreciate it.

/A ←s→ B might very well be true, but you can't infer it just from A ←s→ /B.

/A and A are just two sets that don't intersect. They also could just be called △and ☐, △ including everything that is not a ☐.

I agree. I was finally able to wrap my head around it. Thanks again for taking time from your day to respond to me.