A note about "if someone...."

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In class, as an exercise we translated and proved the following:

Anyone in the group who goes to London will go to Paris. Jones and Smith are in the group. If someone goes to London, Smith will. Jones will go to London. Therefore, Smith will go to Paris.

Turning this into English that's on the way to our logic, we could make this:

Anyone in the group who goes to London will go to Paris.
Jones is in the group and Smith is in the group.
If someone goes to London, Smith will go to London.
Jones will go to London.
Therefore, Smith will go to Paris.

And then in our logic:

/\x((Fx^Gx)-->Hx)
(FA ^ FB)
\/xGx --> GB
GA
CONCLUSION: HB

Now, I can't draw boxes in HTML, but since this is a direct proof with no subproof we should be alright. Here's the proof (cross out the "show", and put a box around all the other lines):

1. Show HB
2. /\x((Fx^Gx)-->Hx)
3. (FA ^ FB)
4. \/xGx --> GB
5. GA
6. \/xGx EG5
7. GB MP 4, 6
8. FB S3
9. FB ^ GB Adj 8, 7
10. (FB ^ GB) --> HB UI 2
11. HB MP 9, 10

But I had hesitated over line 4. It can in fact be translated in two equivalent ways. I hesitated because I can't think of a way to make this intuitive; the translations are:

/\x(Gx --> GB)

Now, this equivalence is actually a theorem in our book. It is theorem 221:

T221 /\x(Fx --> P) <--> (\/xFx --> P)

So, an alternative translation would be:

/\x((Fx^Gx)-->Hx)
(FA ^ FB)
/\x(Gx --> GB)
GA
CONCLUSION: HB

And the alternative proof would be:

1. Show HB
2. /\x((Fx^Gx)-->Hx)
3. (FA ^ FB)
4. /\x(Gx --> GB)
5. GA
6. GA --> GB UI4
7. GB MP 5, 6
8. FB S3
9. FB ^ GB Adj 8, 7
10. (FB ^ GB) --> HB UI 2
11. HB MP 9, 10

Hope that helps those of you who wonder about why I hesitated over the translation and whether to even mention this issue.