SD+ +-----------------------------------------------------------------------------+ | PREMISE | 1| ¬A v (B & C) | | | PREMISE | 2| ¬D -> ¬B | | | PREMISE | 3| ¬(¬A v H) | | | DM3 | 4| ¬¬A & ¬H | | | &EL4 | 5| ¬¬A | | | &ER4 | 6| ¬H | | | DS5,1 | 7| B & C | | | &EL7 | 8| B | | | DN8 | 9| ¬¬B | | | MT2,9 | 10| ¬¬D | | | DN10 | 11| D | | | &I11,6 | 12| D & ¬H | | +-----------------------------------------------------------------------------+

+------------------------------------------ | 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 An alternate version (doing vE on the outside) is +------------------------------------------ | PREM | 1 | ¬ A v B | ASS | 2 | +-- ¬A | ASS | 3 | | +-- A | ASS | 4 | | | +-- ¬B | R 3 | 5 | | | | A | R 2 | 6 | | | +-- ¬A | ¬E 4-5,4-6 | 7 | | +-- B | ->I 2-7 | 8 | +- A -> B | ASS | 9 | +-- B | ASS | 10 | | +- A | R8 | 11 | | +-- B | ->I 2-7 | 12 | +- A -> B | vE 1,2-7,8-9 | 13 | A -> B +------------------------------------------ No (more) errors

- from the top - we need to use vE to use the Premise.
- from the bottom - it seems like a usefull stratey to use ->I to get the final line.

- vE on the outside (second version) with -> I inside each 'case';
- ->I on the outside (first version) with vE on the inside to use the premise.

Give an English version for this same problem.

Suppose ¬ A v B is true.

There are now two cases: ¬ A is true * or * B is true.

Case I: Assume ¬ A is true. Then A is false.

When A is false, then A->B is always true.

Case II: Assume B is true. Then A->B is true.

In both cases, we see that A-> B is true, as required.

(a) Premise (A v B) -> C ______________ Conclusion B -> CNote. You can convince yourself (even me) that this is valid by using truth tables or by using the test for derivability.Valid+------------------------------------------ | PREM | 1 | A v B -> C | ASS | 2 | +-- B | vI 2 | 3 | | A v B | ->E 1,3 | 4 | +-- C | ->I 2-4 | 5 | B -> C +------------------------------------------ No (more) errors

However, the question clearly asked you for a derivation, so you will not get many marks unless you give a correct derivation.

I was suprised at the number of people who gave a correct, but complicated, SD+ derivation. Notice that the above derivation is short and entirely in SD! I think it really captures the logical process in a much clearer way.

(b) Premise B -> C ______________ Conclusion (A v B) -> CThis tva makes the premise true and the conclusion false. Therefore the argument isInvalidtruth value assignment: A B C | B -> C | ( A v B ) -> C ----------|-------------|------------------ T F F | F T F | T T F T F

Note. This tva can been found by working backwards.
To show that this is invalid, we must find a tva making

B -> C True and ( A v B ) -> C False.

(A v B) -> C False requires C to be False and (A v B) to be True.

However, C False and B -> C true, requires B to be False.

Finally, B false and (A v B) True, requires A to be true.

We conclude that the required tva is T F F.

**Remarks**When faced with this choice, I first
explore the options:

- make the conclusion false;
- make the premise(s) all true.

- Soundness:

If \Gamma |-^{SD}\beta then \Gamma |=_SL \beta. - Completeness:

If \Gamma |=_SL \beta then \Gamma |-_SD \beta.

(a) If \Gamma |-SD+ \alpha -> \beta then \Gamma |-SD+ \beta.

** False **

[Hopefully, you did * not * find an SD+ derivation connecting
the antecedent: [\Gamma |-SD+ \alpha -> \beta ] and the consquent: [\Gamma |-SD+ \beta]. If you did, you made a serious logical error.!]

The counter example could be as simple as:

- \Gamma = { }; \alpha = A, \beta = A;
- It is clear that A -> A has an SD+ derivation from no premises:
+------------------------------------------ | ASS | 1 | +-- A | ->E 1,3 | 2 | +-- A | ->I 2-4 | 3 | A -> A +------------------------------------------

- However, A does
*not*have a derivation from no premises.

The truth value assignment A = F satifies the empty set, but does not satisfy the conclusion: A. - This example with the antecendent [\Gamma |-SD+ \alpha -> \beta ] true and the conseqent [\Gamma |-SD+ \beta] false shows the sentence if false (at least sometimes).

Note: Of course, there are examples where the sentence is true. That does not matter.

An answer **True** means that you claim the sentence is true for * all * choices of \Gamma, \alpha and \beta, including the choices I may make in by perversity!
I choose the example above!

(b) If [ \Gamma U {¬ \beta} is SD-consistent ] then not [ \Gamma |= \beta ].

** True **

- First translate the sentence into clear words (and logical connectives):

If not [ \Gamma U {¬ \beta} is SD-inconsistent ]

then not [ \Gamma |= \beta ]. - This form suggests we look at the (logically equivalent) contrapositive:

If [ \Gamma |= \beta ] then [ \Gamma U {\neg \beta} is SD-inconsistent ]. - We should now consider putting both sides into the same vocabulary:
either both in terms of SD derivations or both int terms of truth value assignments. I will choose to put it all in terms of derivations here (but the other choice also works - see below).

By completeness: [ \Gamma |= \beta ] implies [ \Gamma |- \beta ]. - Our problem is now of the form:

If [ \Gamma |- \beta ]. then [\Gamma U {¬ \beta} |- alpha and \Gamma U {¬ \beta} |- ¬\alpha. for some \alpha ]. - I can now see the connection between one derivation and the other:

Suppose [ \Gamma |- \beta ]:Premises \Gamma .... ...... justifications wffs ." " justification \beta If we add ¬\beta as a premise, we have a new expanded derivation: Premises \Gamma Premise ¬\beta .... ...... justifications wffs ." " justification \beta R ¬\beta

This second derivation is precisely the demonstration that \Gamma U {¬ \beta} is SD-inconsitent.