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.
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 ☐.
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.
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4 comments
@dimakyure869 said:
@businesskarafa858 said:
@dimakyure869 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.
@businesskarafa858 said:
@dimakyure869 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 ☐.
@dimakyure869 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.
edit... it looks like you are saying this?
A ←s→ /B
B ←s→ /A
In which case, no they are not.