- 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. - 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.)

- 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. - 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.

- 5.4.7 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

- 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. (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. - 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. - 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. - 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.