Solutions for Assignemnt 2

Question #1.

3.1.6.

The translations are

a. ~W -> ~B
b. ~(B -> W)
c. W v ~W
d. ~(W&~B)
e. ~W v B
f. B -> W
g. B -> B
h. ~(~W -> ~B)

a is logically equivalent to f;
b is logically equivalent to h;
c is logically equivalent to g;
d is logically equivalent to e.

Question #2


 TRUTH-TABLE : 333H                                                            
     +----------------------------------------------------------------------+  
     |                                                                      |  
     |----------------------------------------------------------------------|  
     |DHRS|H, ~(RvS)=~H, (~D&~S)&~R                                         |  
+----+----+-----------------------------------------------------------------+-+
| | 1|TTTT|T  F TTT TFT   FTFFT FFT                                         | |
| | 2|TTTF|T  F TTF TFT   FTFTF FFT                                         | |
| | 3|TTFT|T  F FTT TFT   FTFFT FTF                                         | |
| | 4|TTFF|T  T FFF FFT   FTFTF FTF                                         | |
| | 5|TFTT|F  F TTT FTF   FTFFT FFT                                         | |
| | 6|TFTF|F  F TTF FTF   FTFTF FFT                                         | |
| | 7|TFFT|F  F FTT FTF   FTFFT FTF                                         | |
| | 8|TFFF|F  T FFF TTF   FTFTF FTF                                         | |
| | 9|FTTT|T  F TTT TFT   TFFFT FFT                                         | |
| |10|FTTF|T  F TTF TFT   TFTTF FFT                                         | |
| |11|FTFT|T  F FTT TFT   TFFFT FTF                                         | |
| |12|FTFF|T  T FFF FFT   TFTTF TTF                                         | |
| |13|FFTT|F  F TTT FTF   TFFFT FFT                                         | |
| |14|FFTF|F  F TTF FTF   TFTTF FFT                                         | |
| |15|FFFT|F  F FTT FTF   TFFFT FTF                                         | |
| |16|FFFF|F  T FFF TTF   TFTTF TTF                                         | |
+-----------------------------------------------------------------------------+
      0000000001111111111222222222233333333334444444444555555555566666666667
      1234567890123456789012345678901234567890123456789012345678901234567890

No (more) errors

From the truth table above, we see that no truth-value assignment satisfies all sentences in the set. Therefore,

Ans: The set is (TRUTH-FUNCTIONALLY) INCONSISTENT.

Informal argument: Suppose the set is consistent; then, for some tva , each sentence gets value T. In particular, (~D&~S)&~R gets value T, hence R and S get value F, and therefore, ~(RvS) gets value T. However, H must get value T; hence ~H becomes F. This implies that ~(RvS)=~H gets value F, which contradicts the assumption that each sentence gets value T. So there is no such t.v.a., so the set is inconsistent.

Question #3:

3.4.4 (b)

 TRUTH-TABLE : 344B                                                            
     +----------------------------------------------------------------------+  
     |                                                                      |  
     |----------------------------------------------------------------------|  
     |P   |Pv~P ,  P                                                        |  
+----+----+-----------------------------------------------------------------+-+
| | 1|T   |TTFT    T                                                        | |
| | 2|F   |FTTF    F                                                        | |
+-----------------------------------------------------------------------------+
      0000000001111111111222222222233333333334444444444555555555566666666667
      1234567890123456789012345678901234567890123456789012345678901234567890

No (more) errors

From line 2 in the truth table above, we see that when P is assigned the value F , Pv~P becomes T while P has value F. Therefore,

Ans: the argument is not (truth-functionally) valid.

3.4.4 (f)


 TRUTH-TABLE : 344F                                                            
     +----------------------------------------------------------------------+  
     |                                                                      |  
     |----------------------------------------------------------------------|  
     |AB  |~(A&B),  ~Av~B                                                   |  
+----+----+-----------------------------------------------------------------+-+
| | 1|TT  |F TTT    FTFFT                                                   | |
| | 2|TF  |T TFF    FTTTF                                                   | |
| | 3|FT  |T FFT    TFTFT                                                   | |
| | 4|FF  |T FFF    TFTTF                                                   | |
+-----------------------------------------------------------------------------+
      0000000001111111111222222222233333333334444444444555555555566666666667
      1234567890123456789012345678901234567890123456789012345678901234567890

No (more) errors

From the truth table above, we see that whenever ~(A&B) has value T, so does ~Av~B. Therefore,

Ans: The argument is (truth-functionally) valid.

Here is an informal argument for 5.4.6b.: (The modified truth table has 16 lines. (And the original would have had 64 lines!!))

Assume the three premises (K -> J) -> R, ¬R v D, (¬D & M) & S are true.
Since (¬D & M) &S is true, we conclude ¬D is true.
Since ¬D is true, D is false.
Since D is false and ¬R v D is true, we conclude that ¬R is true.
Since ¬R is true, R is false.
Since R is false and (K -> J) -> R is true, we conclude (K -> J) is false.
Therefore the conclusion of the argument, ¬(K -> J), is true.

Since our assumption that the premises are true leads to the fact that the conclusion is true,
the argument is valid.

Question #4:

3.4.8 (b):

If \alpha and \beta are truth-functionally indeterminate then {\alpha, \beta} is truth-functionally consistent.

Ans: false: Counterexample: Let \alpha be A and \beta be ~A.
This must be about the simplest possible counterexample.

3.4.8 (d):

Any argument with an inconsistent set of premises is logically valid.

Ans: true.

Proof: Suppose the argument is invalid. Let \Gamma be the set of premises, and \alpha be the conclusion. Then there must be a tva assignment \tau such that each sentence in \Gamma gets value T under the tva, while \alpha gets value F.

However, the existence of such a \tau indicates that \Gamma is consistent. This contradicts the assumption that \Gamma is inconsistent.