advanced formal logic some and most inference

cynthia.wu82

Hi guys,
I am confused with a logic inference.
According to a book, if A, then not B, can be notated like A-->/B or A<-|->B. However, in 7sage, it is maybe notated like A<--->/B.

Here's the question.
1. E<-|->F--> G -->H (the original one)

According to the question key, the inference is,
G<--s-->/E
F-->H
H<--s-->/E

But, if we use E-->/F to replace the part of E<-|->F (because according to the very beginning of the post,A-->/B AND A<-|->B are actually the same thing ), we get,

1. E-->/F--> G -->H (the replaced first )
   then the inference are,
   E-->G-->H (this is very difference with the original above, which is not inference can be made like this)
   E-->H (Which according to the original above, there is not such inference can be made like this)
   /F-->H (which is different from the above original inference which is F-->H)

However, if we replace the original part of E<-|->F with E<--->/F, interestingly, the inference different with above two.

Here we have:

E<--->/F--> G -->H
and the inference we have are
except we can make the exactly same inference with the 2 ones, we also can infer that,
G<--s-->/F
G<--s-->E
E<--s-->H

these inferences are totally different with the original ones.

the trick thing is according to the book, E<-|->F is the same with E-->/F, and according to 7sage, actually E<-|->F is E<--->/F, which leading to the hypothesis that if we replace the E<-|->F to whatever these two different versions, the inferences made should be the same. Who can please clarify me?

Thank you!

Cynthia

Sami

@"cynthia.wu82" said:
Hi guys,
I am confused with a logic inference.
According to a book, if A, then not B, can be notated like A-->/B or A<-|->B. However, in 7sage, it is maybe notated like A<--->/B.

Here's the question.
1. E<-|->F--> G -->H (the original one)

According to the question key, the inference is,
G<--s-->/E
F-->H
H<--s-->/E

But, if we use E-->/F to replace the part of E<-|->F (because according to the very beginning of the post,A-->/B AND A<-|->B are actually the same thing ), we get,

1. E-->/F--> G -->H (the replaced first )
   then the inference are,
   E-->G-->H (this is very difference with the original above, which is not inference can be made like this)

I think over here you are making the mistake of equating F--->G
and linking it in this chain with /F--->G
In the first chain All things F are G
In the second sentence you have linked up all things "not F" are G.
They mean two different things.

If you write it according to the first sentence:
F-->G-->H
and if you link it up with E--->/F you will get:
F--->G----->H and F--->/E (contrapositive of the above sentence helps me to see the inference better)

if all of F is G and all of F is /E, then we know there is at lease a some intersection between G and E. (G <-s->/E)
and if all of F is H and all of F is /E, the again there is a some intersection between H and /E. (H <-s-> /E)

*These statements are equivalent to the inferences you derived in your first example.
I hope this helps.

cynthia.wu82

my question is whether E<-|->F, E<--->/F, and E--->/F are the same thing?
in other words, should the" if A, then not B" be biconditional notated?

Heart Shaped Box

No, it's not a Bi-conditional. If A then not B is a not both rule: A ---> B, for it's not clear if it's required for A to even show up. "If...then..." is a conditional itself. The meaning of this statement does not equate a bi-conditional.

Sami gave you a much more detailed breakdown of the inferences and possible errors made due to the confusion on the concept. It does a much better job of explaining the nuances on the logic.