1. 5.3.3 b. (It is very hard to type this in html....)

 SD : 533B
+----------------------------------------------
| PREM              |  1 |  ¬ P -> R
| PREM              |  2 |  ¬ R
| ASS               |  3 |  +- P
| ASS               |  4 |  | +- S
| R3                |  5 |  | +- P
| ->I 4-5           |  6 |  +- S -> P
| ASS               |  7 |  +- S -> P
| ASS               |  8 |  | +- ¬ P
| ->E 1,8           |  9 |  | |   R
| R2                | 10 |  | +- ¬ R
| ¬E 8-9,8-10  | 11 |  +- P
| \bicondI 3-6,7-11 | 12 |  P \bicond (S -> P)
+-----------------------------------------------

No (more) errors 
5.3.3 (b) (Whiteley's section only) By now, you will have realized that this is the same as part (a) of the final problem - with an extra 'wrapper' formed by a subderivation (with the premise of (a) becoming the Assumption of this problem) and an application of the rule >I to discharge the assumption.

2. 5.4.1 b.
False. Let \alpha be the wff G & ¬ G . Then \alpha is a contradiction, so {\alpha} is an inconsistent set, hence it entails EVERY wff \beta (in particular, ¬ \alpha).

(Proof: yet again.... *sigh* ... There is no t.v.a. \tau such that \tau |= {\alpha}. So there is no \tau such that:

\tau satisfies {\alpha} and \tau does NOT satisfy \beta.

So every \tau that satisfies {\alpha} also satisfies \beta.) (There was never a time I played Fischer when I did not crush him.)

3. 5.4.4 f.
  SD : 544F
+-----------------------------------------------------
| PREM                  |  1 |     A \bicond ¬¬ B
| PREM                  |  2 |     B -> ¬ M
| PREM                  |  3 |     (A v B) & M
| &ER 3                 |  4 |     M
| &EL 3                 |  5 |     A v B
| ASS                   |  6 |  +-- A
| \bicondE 1,6          |  7 |  |  ¬¬ B
| ASS                   |  8 |  | +--¬ B
| R8                    |  9 |  | |  ¬ B
| R7                    | 10 |  | +--¬¬ B
| ¬E 8-9,8-10           | 11 |  +-- B
| ASS                   | 12 |  +-- B
| R12                   | 13 |  +-- B
| vE 5,6-11,12-13       | 15 |     B
| ->E 2,15              | 16 |     ¬ M
+--------------------------------------------------------

No (more) errors 
Our pair of contradictory wff's appears in lines 4 and 16, showing that the set consisting of our premises is SD-inconsistent.
4. 5.4.5 d.
 SD : 545D
+------------------------------------------
| PREM                |  1 |     D -> M
| PREM                |  2 |     R \bicond (¬M & S)
| PREM                |  3 |     R
| PREM                |  4 |     ¬¬ D
| \bicondE 2,3        |  5 |     ¬M & S
| &EL 5               |  6 |    ¬M
| ASS                 |  7 |  +-- ¬ D
| R7                  |  8 |  |   ¬ D
| R4                  |  9 |  +-- ¬ D
| ¬E 7-8,7-9     | 10 |     D
| ->E 1,10            | 11 |     M
+------------------------------------------

No (more) errors
Lines 6, 11 show that our test set is SD-inconsistent, as the
test for derivability requires.

5. 5.4.7 h.
 SD : 547H
+-----------------------------------------------------
| PREMISE               |  1 |     (A & B) v (A & C)
| ASSUMPTION            |  2 |  +--  A & B
| &ER 2                 |  3 |  |    B
| &EL 2                 |  4 |  |    A
| vIR 3                 |  5 |  |    B v C
| &I 4,5                |  6 |  +--  A & (B v C)
| ASS                   |  7 |  +--  A & C
| &EL 7                 |  8 |  |    A
| &ER 7                 |  9 |  |    C
| vIL 9                 | 10 |  |    B v C
| &I 8,10               | 11 |  +--  A & (B v C)
| vE 1,2-6,7-11         | 12 |       A & (B v C)

| PREMISE               | 314 |     A & (B v C)
| &EL 314               | 315 |     A
| &ER 314               | 316 |     B v C
| ASS                   | 317 |  +-- B
| &I 315,317            | 318 |  |   A & B
| vIR 318               | 319 |  +-- (A & B) v (A & C)
| ASS                   | 320 |  +-- C
| &I 315,320            | 321 |  |   A & C
| vIL 321               | 322 |  +-- (A & B) v (A & C)
| vE 316,317-319,320-322| 323 |     (A & B) v (A & C)
+---------------------------------------------------

No (more) errors 
6. 7. (a)

SD : 7A
+-------------------------------------------
| PREM                  |  1 |     A -> B
| ASS                   |  2 | +--¬ B
| ASS                   |  3 | | +-- A
| R 2                   |  4 | | |   ¬ B
| ->E 1,3               |  5 | | +-- B
| ¬I 3-4,3-5            |  6 | +-- ¬ A
| ->I 2-6               |  7 |     ¬ B -> ¬ A
+-------------------------------------------

No (more) errors

English Version: (Ganong)

First, some remarks:
In these English versions, we have referred to
truth-value assignments making the premises true. You do not have to
do this when you write down English versions of derivations. But you
should be aware that when you write something like
"Suppose C -> B is true; then we have ....",

what you really mean is,

"Suppose that we have a t.v.a. for which the wff
C -> B is true. Then for that t.v.a., we have ...."

Another remark: Whiteley's "English versions" seem to me to be,
really, efficient accounts of why we believe each argument to be
valid. My "English versions" are, rather, attempts to verbalize
(ugly word, that) the derivations, in the sort of English a
mathematician would use. I will not ask my vict -- er, students to
produce English versions like the ones I have typed here. But it may
do some good for some people to see my versions here; same for
Whiteley's. He is right that people I know who think logically think
in the bigger "chunks" afforded by SD+, rather than in the choppy
little steps of SD.

OK. Now for the English version of 7. (a):

Suppose A -> B "is true" (i.e., has value "true"
under some truth value assignment which we keep fixed throughout the
argument). And assume ¬ B is true. Then we claim ¬ A must be
true. For otherwise, A would be true, hence, by our premise, B is
true. But we assumed ¬ B is true. So ¬ A is true. We have shown
that given our premise, whenever ¬ B is true, ¬ A is also true.
So given our premise, ¬ B -> ¬ A is true.

English Version  (Whiteley):

Suppose  A -> B is true.

Case I If  A  is true, then  B is true and  ¬B  is false.
If  ¬B  is false  then  ¬B -> ¬A  is
automatically true.
Case II  Otherwise  ¬A is false.
Therefore ¬B -> ¬A  is true.

In all cases we conclude that ¬B -> ¬A  is true.

7. 7. (b)

SD : 7B
+------------------------------------------
| PREM                  |  1 |   ¬ A v B
| ASS                   |  2 | +-- A
| ASS                   |  3 | | +-- ¬A
| ASS                   |  4 | | | +-- ¬B
| R 2                   |  5 | | | |   A
| R 3                   |  6 | | | +-- ¬A
| ¬E 4-5,4-6            |  7 | | +-- B
| ASS                   |  8 | | +-- B
| R8                    |  9 | | +-- B
| vE 1,3-7,8-9          | 10 | +-- B
| ->I 2-10              | 11 |    A -> B
+------------------------------------------

No (more) errors

English Version  (Ganong):

Suppose ¬ A v B "is true" (see English version of 7. (a)).
And assume that A is true. Now, either ¬ A is true or B is true,
by our supposition at the beginning (our premise). We will show that
in either case, B must be true.

So suppose ¬ A is true. (Note that now we have a contradiction
already, with the fact that A is true, so we can prove anything we
want.) Then ¬ B could not be true, for if it were then A and ¬ A
would be true, by earlier assumptions. So B must be true.

If, on the other hand, B is true, then B is true. (Argue with THAT!)

So B is true (since we have shown it is true in both cases).

English Version  (Whiteley):
Suppose  ¬A v B is true.
Therefore either ¬A is true or
B is true.
There are now two cases:

Case I: ¬A is true.
Therefore A is false.
Therefore  A -> B is true.
Case II:  B is true.  Therefore  A -> B is true.

Since both cases lead to  A -> B  being true
and there are only two cases,
we conclude that  A ->B  is true.

Note  This is not a direct translation of the
derivation.  Nor is the derivation a direct translation of the
English.  However, they  are  related at a fundamental
level.  The English is much closer to the possible SD+ derivation.

8. 7. (c)
 SD : 7C
+---------------------------------------------
| ASS                   |  1 | +-- ¬ (A v ¬ A)
| ASS                   |  2 | | +-- A
| vIR 2                 |  3 | | |   A v ¬ A
| R 1                   |  4 | | +-- ¬ (A v ¬ A)
| ¬I 2-3,2-4       |  5 | |    ¬ A
| vIL 5                 |  6 | |    A v ¬ A
| R 1                   |  7 | +-- ¬ (A v ¬ A)
| ¬E 1-6,1-7        |  8 |     A v ¬ A
+---------------------------------------------
No (more) errors

English Version  (Ganong):

We claim that A v ¬ A is true under any t.v.a.
So suppose \tau is
a t.v.a., which we keep fixed throughout this argument.

Claim: It is not
possible that the wff in line 1 above is made true by \tau. For, suppose
it were: Then we claim that ¬ A would be true. For, if A were true,
then A or ¬A would be true, hence A v ¬A would be true. But
this contradicts our supposition that the wff in line 1 is true. So
¬A is true.

But then A is true or ¬ A is true, i.e., A v ¬ A
is true. But we assumed that its negation was true. Contradiction. So
the wff in line 1 is not true under \tau. So the wff in line 8 is.
(And since we never said what \tau was, this holds for
arbitrary \tau.)

English Version  (Whiteley):

Suppose  A -> B is true.

Case I   If  A  is true, then  B is true and  ¬B  is false.
If  ¬B  is false  then  ¬B -> ¬A  is
automatically true.
Case II  Otherwise  ¬A is false.
Therefore ¬B -> ¬A  is true.

In all cases we conclude that ¬B -> ¬A  is true.

9. 7. (b)

SD : 7B
+------------------------------------------
| PREM                  |  1 |   ¬ A v B
| ASS                   |  2 | +-- A
| ASS                   |  3 | | +-- ¬A
| ASS                   |  4 | | | +-- ¬B
| R 2                   |  5 | | | |   A
| R 3                   |  6 | | | +-- ¬A
| ¬E 4-5,4-6            |  7 | | +-- B
| ASS                   |  8 | | +-- B
| R8                    |  9 | | +-- B
| vE 1,3-7,8-9          | 10 | +-- B
| ->I 2-10              | 11 |    A -> B
+------------------------------------------

No (more) errors

English Version  (Ganong):

Suppose ¬ A v B "is true" (see English version of 7. (a)).
And assume that A is true. Now, either ¬ A is true or B is true,
by our supposition at the beginning (our premise). We will show that
in either case, B must be true.

So suppose ¬ A is true. (Note that now we have a contradiction
already, with the fact that A is true, so we can prove anything we
want.) Then ¬ B could not be true, for if it were then A and ¬ A
would be true, by earlier assumptions. So B must be true.

If, on the other hand, B is true, then B is true. (Argue with THAT!)

So B is true (since we have shown it is true in both cases).

English Version  (Whiteley):
Suppose  ¬A v B is true.  Therefore either ¬A is true or B is true.
There are now two cases:

Case I: ¬A is true.
Therefore A is false.
Therefore  A -> B is true.
Case II:  B is true.  Therefore  A -> B is true.

Since both cases lead to A->B being true and there are only two cases,
We conclude that  A->B  is true.

Note  This is not a direct translation of the
derivation.  Nor is the derivation a direct translation of the
English.  However, they  are  related at a fundamental
level.  The English is much closer to the possible SD+ derivation.

10. 7. (c)
 SD : 7C
+---------------------------------------------
| ASS                   |  1 | +-- ¬ (A v ¬ A)
| ASS                   |  2 | | +-- A
| vIR 2                 |  3 | | |   A v ¬ A
| R 1                   |  4 | | +-- ¬ (A v ¬ A)
| ¬I 2-3,2-4            |  5 | |    ¬ A
| vIL 5                 |  6 | |    A v ¬ A
| R 1                   |  7 | +-- ¬ (A v ¬ A)
| ¬E 1-6,1-7              |  8 |     A v ¬ A
+---------------------------------------------
No (more) errors

English Version  (Ganong):

We claim that A v ¬ A is true under any t.v.a.
So suppose \tau is
a t.v.a., which we keep fixed throughout this argument.

Claim: It is not
possible that the wff in line 1 above is made true by \tau. For, suppose
it were: Then we claim that ¬ A would be true. For, if A were true,
then A or ¬A would be true, hence A v ¬A would be true. But
this contradicts our supposition that the wff in line 1 is true. So
¬A is true.

But then A is true or ¬ A is true, i.e., A v ¬ A
is true. But we assumed that its negation was true. Contradiction. So
the wff in line 1 is not true under \tau. So the wff in line 8 is.
(And since we never said what \tau was, this holds for
arbitrary \tau.)

English Version  (Whiteley):

There are two possibilities:  A  is true  or  A  is false.

Case I   If  A  is true, then A v ¬A  is true.
Case II  Otherwise  A  is false and  ¬A  is true.
Therefore  A v ¬A  is true.

In all cases we conclude that A v ¬A  is true.