## Solutions to Whiteley's Test I

### Question 1: 12 marks

 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                                    | |
+-----------------------------------------------------------------------------+



### Question 2: 10 marks

This question is right off assignment 3! I repeat the solution given there

+------------------------------------------
| 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

Note The basic strategy here is a mix of two parts:
• 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.
Both of these strategies require subderivations with specific Assumptions. The only choice is what order to nest the two subderivations in:
• 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.
Notice below that the English version I give corresponds to the first (longer) alternative, with the extra benifit of the informal 'rules of SD+' which we regualary use in our informal reasoning.

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.

### Question 3: 8 marks

Valid or Invalid? In each case, if the argument is valid, give an SD+ derivation and if the argument is invalid, give a line from the truth table which demonstrates it is invalid.

(a)  Premise     (A v B) -> C
______________
Conclusion      B -> C

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

Note. You can convince yourself (even me) that this is valid by using truth tables or by using the test for derivability.
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) -> C

Invalid
truth value assignment:

A  B  C  |   B  ->  C  |  ( A v B )  ->  C
----------|-------------|------------------
T  F  F  |    F  T  F  |   T  T  F  T   F

This tva makes the premise true and the conclusion false. Therefore the argument is not valid.

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.

RemarksWhen faced with this choice, I first explore the options:

• Inspect for the possibililty of a quick derivation.
• Look at the meaning, perhaps even translate into English to see if it 'sounds right'.
• If the first two do not make you 'believe' it is true, then I would seriously consider the possibility it is false.
• If I consider the possibility that it is invalid, then I specifically look for the lines of the truth table which:
• make the conclusion false;
• make the premise(s) all true.
• It is possible that I find it very hard' (impossible?) to make this line which demonstrates the argument is invalid. This is convincing evidence that the argument is valid. In fact, this 'impossibility' may actually suggest a specific 'proof by contradiction' for the desired outcome. (This is proactical form of the 'test of derivability'!)

### Question 4: 2 marks

Give exact statements of the Soundness Theorem and the Completeness Theorem. [You may use symbols in these statements.]
• Soundness:
If \Gamma |-SD \beta then \Gamma |=_SL \beta.
• Completeness:
If \Gamma |=_SL \beta then \Gamma |-_SD \beta.

### Question 5: 8 marks

True or False? If False give a counterexample. If True give your reasoning. You may use the Soundness or Completeness Theorem (for SD and for SD+) - but identify which you are using.

(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: It is note possible to give a clear counterexample unless you specify \Gamma. If you leave \Gamma unspecified, than I will make my own choice. I choose \Gamma = {A & ¬A}. WIth this choice, the sentence is true, regardless of what you pick for \alpha and \beta. The only way to stop this silly choice is for you to make a better choice before I have a chance to interfere!

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.
Note: Some students recognized this as directly related to the 'test for derivability'. It is - but you still need to give an explicit argument about the connections. In particular, you cannot switch from SD-inconsistent to truth functionally consistent without speaking about soundness etc.