Formal Logic Additive Inference Drill

[HarmonRabb]

Hi,
In the drill's first problem, I came up with more additive inferences than the answer key listed. Can you explain to me where I went wrong:

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All Es are Fs
Some Fs are Gs
No Gs are Hs

I correctly came up with the initial diagram of:

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E--->F<---s--->G<---|--->H

My inferences:

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F<---s--->/H
E<---s--->/H 
E<---s--->G

From my list, only the first one was in the answer key.

I obtained E<---s--->/H by recycling F<---s--->/H and thus I get E--->F<---s--->/H and thus E--->/H

I obtained E<---s--->G by following the all arrow from E to F and the some train from F to G.

[Dave Killoran]

The problem occurs at: E--->F<---s--->/H. There is no inference from that relationship.

Broadly put, the way I think about it is that the problem occurs in the direction of the arrow as it relates to Some. Arrow leading away from Some gives you an inferences, arrow leading into Some typically won't (I say typically because things like double arrows exist, but in the single arrow world, this is the case). This broad rules means the following:

A B C
Arrow leads "away" from the Some, there's an inference: A C

A B C
Arrow points "at" the Some, no inference.

Once you start seeing the connections in a visual way like that, it becomes easier to see the inferences super fast