• 5.5.3 f

 SD+ : 553F
+----------------------------------------------------------------------------+
| ASSM                  |  1 | +-- A
| DN1                   |  2 | +-- ¬¬A
| ASSM                  |  3 | +-- ¬¬A
| DN3                   |  4 | +-- A
| \bicondI 1-2,3-4      |  5 |   A \bicond ¬¬A
+-----------------------------------------------------------------------------+

Informal Argument (Whiteley).
Assume A is true. Then ¬A is false and ¬¬A is true.
Conversely, assume ¬¬A is true. Then ¬A is false and A is true
We conclude that the biconditional holds. (Ganong: I agree that this English shows the validity of the argument.)

• 5.5.3 h
 SD+ : 553H
+-----------------------------------------------------------------------------+
| PREMISE               |  1|     ¬(P & ¬Q)                                 | |
| PREMISE               |  2|     ¬Q v R                                    | |
| PREMISE               |  3|     S -> ¬R                                   | |
| ASSUMPTION            |  4|+----P                                         | |
| DM1                   |  5||    ¬P v ¬¬Q                                  | |
| DN4                   |  6||    ¬¬P                                       | |
| DS5,6                 |  7||    ¬¬Q                                       | |
| DS2,7                 |  8||    R                                         | |
| DN8                   |  9||    ¬¬R                                       | |
| MT3,9                 | 10||    ¬S                                        | |
| &I10,9                | 11||    ¬S & ¬¬R                                  | |
| DM11                  | 12|+----¬(S v ¬R)                                 | |
| ->I4-12               | 13|     P->¬(Sv¬R)                                | |
+-----------------------------------------------------------------------------+

• 5.5.4 h
 SD+ : 554H
+-----------------------------------------------------------------------------+
| PRE                   |  1|     H v G                                     | |
| PRE                   |  2|     (K v ¬I) & (¬H v K)                       | |
| PRE                   |  3|     (¬K & I) v ¬(H v G)                       | |
| DN1                   |  4|     ¬¬(H v G)                                 | |
| DS3,4                 |  5|     ¬K & I                                    | |
| &EL2                  |  6|     K v ¬I                                    | |
| DN6                   |  7|     ¬¬(K v ¬I)                                 | |
| DM7                   |  8|     ¬(¬K & ¬¬I)                               | |
| DN8                   |  9|     ¬(¬K & I)                                 | |
+-----------------------------------------------------------------------------+

• 5.5.5 b

The test set is { ¬E v N , S = (¬N & A), ¬¬S , ¬¬E }. The following derivation shows that the test set is SD+-inconsistent. Therefore, ¬E is derivable from the given set.

 SD+ : 555B
+-----------------------------------------------------------------------------+
| PREMISE               |  1|     ¬E v N                                    | |
| PREMISE               |  2|     S = (¬N & A)                              | |
| PREMISE               |  3|     ¬¬S                                       | |
| PREMISE               |  4|     ¬¬E                                       | |
| DN3                   |  5|     S                                         | |
| =E2,5                 |  6|     ¬N & A                                    | |
| &EL6                  |  7|     ¬N                                        | |
| DS1,7                 |  8|     ¬E               ( \alpha )               | |
| R4                    |  9|     ¬¬E              (¬ \alpha )              | |
+-----------------------------------------------------------------------------+

• 5.5.6 h

 SD+ : 556H
+-----------------------------------------------------------------------------+
| PRE                   |  1|     (A & ¬G) v (¬G & S)                       | |
| PRE                   |  2|     (¬S v G) & R                              | |
| &EL2                  |  3|     ¬S v G                                    | |
| DN3                   |  4|     ¬¬(¬S v G)                                | |
| DM4                   |  5|     ¬(¬¬S & ¬G)                               | |
| DN5                   |  6|     ¬(S & ¬G)                                 | |
| CM6                   |  7|     ¬(¬G & S)                                 | |
| DS1,7                 |  8|     A & ¬G                                    | |
| &EL8                  |  9|     A                                         | |
| DN9                   | 10|     ¬¬A                                       | |
| vIL10                 | 11|     ¬¬H v ¬¬A                                 | |
| vIR11                 | 12|     (¬¬H v ¬¬A) v ¬M                          | |
+-----------------------------------------------------------------------------+

Strategy Note that the final conclusion contains brand new letters H and M. Our experience on Assignment 1 indicated that making a new letter have significance is difficult.
There are really two possibilities:
(a) The premises contain a contradiction and we can prove anything (and everything);
(b) The presence of this letter in the conclusion is not essential. This is clearly the case here - H and M are just thrown in at the end (using vI). Even before I do the proof, I recognize that I should be looking for a proof of ¬¬A (line 10).

• 5.5.7 j

 SD+ : 557J
+-----------------------------------------------------------------------------+
| ASSM                  |  1|+----(¬S & ¬A) v (¬¬K & ¬S)                    | |
| CM1                   |  2||    (¬S & ¬A) v (¬S & ¬¬K)                    | |
| DI2                   |  3||    ¬S & (¬A v ¬¬K)                           | |
| DN3                   |  4||    ¬S & (¬A v K)                             | |
| IM4                   |  5||    ¬S & (A -> K)                             | |
| DN5                   |  6||    ¬S & ¬¬(A -> K)                           | |
| DM6                   |  7|+----¬(S v ¬(A -> K))                          | |
| ASSM                  |  8|+----¬(S v ¬(A -> K))                          | |
| DM8                   |  9||    ¬S & ¬¬(A -> K)                           | |
| DN9                   | 10||    ¬S & (A -> K)                             | |
| IM10                  | 11||    ¬S & (¬A v K)                             | |
| DN11                  | 12||    ¬S & (¬A v ¬¬K)                           | |
| DI12                  | 13||    (¬S & ¬A) v (¬S & ¬¬K)                    | |
| CM13                  | 14|+----(¬S & ¬A) v (¬¬K & ¬S)                    | |
| =I1-7,8-14            | 15|     ((¬S & ¬A) v (¬¬K & ¬S)) = ¬(Sv¬(A->K))   | |
+-----------------------------------------------------------------------------+

• Solutions to the True /False questions from 5.6.6.