I.  Assume that the statements below are premises of an argument, though there is no conclusion provided.  Answer each question in abstraction from the others unless a question specifically refers to some other.

1.  All American Philosophers are either logicians or mathematicians.
2.  David is a philosopher.
3.  Bob is not a logician.
4.  There are Americans who are not logicians.
5.  Some philosophers are not Americans.

A.  Is it a proper use of EI to instantiate line 4 to "Ab . ~Lb"?  No, because you can only use EI with an unknown individual, and that has to be a variable, not a constant.
B.  Is it permissible to instantiate line 1 to (Ab . Pb) --> (Lb v Mb)?  Why?   Yes, there are no restrictions on the use of UI.
C.  Is it possible to use conjunction to combine lines 2 and 3 like this?
  (Ex)(Px . ~Lx)?  Why?   No.  These are two separate statements.  You could get (Ex)Px . (Ex)Lx after generalizing each statement individually, but to put them both bound by the same quantifier would indicate that the same person is both a philosopher and not a logician, and that is not warranted by the information given in statements 2 and 3.
D.  If line 5 was instantiated to Px . ~Ax, can you then instantiate line 4 to Ax . ~Lx?  Why?   No, because you have always to instantiate existentially to something previously unused in the proof.
E.  If line 1 was instantiated to (Az . Pz) --> (Lz v Mz), can you then instantiate line 2 to Pz?  Why?   You can't instantiate line 2 to anything because it isn't quantified.
F.  Based on line 3, is it a correct inferential move to assert that ~(x)Lx?  Why? Yes.  if ~Lb, then (Ex)~Lx (by EG), and then ~(x)Lx by QN.
G.  If you know that (x)~Mx, can you then justify the claim that ~Mb?  Why?   Yes, by universal instantiation.
H.  From line 2, can you say that (x)Px?  Why?   No.  David is a known individual, and UI only "works" with arbitrarily selected individuals.
I.  If there existed among the statements one such that ~(Lx v Mx), could it be used with line 1 to produce ~(Ax . Px)?  Why? Yes.  If line 1 is translated as (x)[(Ax . Px) --> (Lx v Mx)], then you use UI to remove the quantifier.  Then use ~(Lx v Mx) against that line with MT.
 

II.  Think of all the statements below as premises of an argument.  The conclusion is irrelevant to the questions that follow, so a specific conclusion is not provided.  The questions following the statements are intended to be answered as either "true" or "false"; in the case of a "false," be able to explain why the statement is false and what the conditions would need to be to make it true.

1.  There are some physicians who are not board certified.
2.  All physicians in XYZ hospital are board certified.
3.  No orderlies are board certified.
4.  Bob is not board certified.
5.  Carol is a physician and Ann is an administrator.
6.  Someone is an orderly.

    Remember that the symbol for the existential quantifier is given in this document as (Ex).

A.  The correct translation of line 1 is: (Ex)(Px . ~Bx)   TRUE
B.  The correct translation of line 2 is: (x)[(Px . Hx) --> Bx]    TRUE
C.  The correct translation of line 3 is: (x)~Bx    FALSE.  The correct translation is (x)(Ox --> ~Bx)
D.  The correct translation of line 4 is: ~Bb   TRUE
E.  The correct translation of line 5 is (Ex)Px . (Ex)Aa    FALSE.  It is Pc . Aa.  There is no quantifier in 5.
F.  You can instantiate line 1 existentially to Pa . ~Ba.   FALSE.  Existential instantiation is limited to unknowns.  "a' is known.
G.  This statement, "(Ex)(Px . Ax)" does not follow from line 5. TRUE
H.  If you instantiate line 1 to Py . ~By, you can use the same variable in the instantiation of line 2.  TRUE, since line 1 is an existential and line 2 is universal.
I.  If you use simplification twice from line 5, you can then conclude oth that ~(x)~Px and ~(y)~Ay.   If you use simp. on line 5 twice, you get Pc on one line, and then Aa on another.  By EG, you get (Ex)Px and by EG, you get (Ey)Ay.  By QN on (Ex)Px, you get ~(x)~Px and by QN on (Ey)Ay, you get ~(y)~Ay.  So this is TRUE.
J.  If you use UI on line 2 to get (Pa . Ha) --> Ba, you can then use EI on line 6 to get Oa.  False.  No, you can't because you can't instantiate existentially to a known individual.
K.  You can conlcude that Oz from line 6 even if another premise stating "(x)Az" was present in the argument. True.
L.  You can conclude that Oa --> ~Ba from line 3 by UG. No, if you take quantifiers off, you are using instantiation, not generalization.
M.  You can instantiate "to anything" from line 4. False.  You can't, because there is no quantifier on line 4.
N.  It follows from line 6 that ~(x)~Ox by QN.   Line 6 says (Ex)Ox.  By QN, that changes to ~(x)~Ox.  So, this is true.
O.  If you instantiate to Py . ~By from line 1, "y" is an arbitrarily selected individual.   True.
P.  Based on line 4, it is correct to infer that ~(x)Bx. If ~Bb, then (Ex)~Bx, and by QN, ~(x)Bx.  True.
Q.  (x)Px follows directly from line 5 by simplification and universal generalization. No, it doesn't.  If Carol is a physician, then you know that Pc.  The individual "c" is known, and it is possible to use universal generalization only from an arbitrarily selected individual.
R.  If you know that (Ex)~Mx, you can justify the claim that ~Mb because it follows by EG.   FALSE.  ~Mb would follow by EI, not EG.
S.  If you were doing a proof based on the premises above, you could instantiate lines 1, 6, and 2, in that order, all to "y".  No, you can't because if you instantiate line 1 to "y" (line 1 is an existential statement), then you can't use "y" again on line 6 (which is also an existential statement), but you could use "y" on line 2 becuase that is a universal statement, and there are no limitations on UI.