QA 34

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course MTH 151

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

015. Conditionals

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Question: `q001. There are 7 questions in this set.

The proposition p -> q is true unless p is true and q is false. Construct the truth table for this proposition.

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Your solution:

p q p->q

T T T

F F F

T F F

F T T

confidence rating #$&*:2

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Given Solution:

The proposition will be true in every case except the one where p is true and q is false, which is the TF case. The truth table therefore reads as follows:

p q p -> q

T T T

T F F

F T T

F F T

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Self-critique (if necessary):

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Question: `q002. Reason out, then construct a truth table for the proposition ~p -> q.

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Your solution:

p q ~p ~p->q

T T F T

F F T F

T F F T

F T T T

confidence rating #$&*:2

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Given Solution:

This proposition will be false in the T -> F case where ~p is true and q is false. Since ~p is true, p must be false so this must be the FT case. The truth table will contain lines for p, q, ~p and ~p -> q. We therefore get

p q ~p ~p -> q

T T F T since (F -> T) is T

T F F T since (F -> F) is T

F T T T since (T -> T) is T

F F T T since (T -> F) is F

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q003. Reason out the truth value of the proposition (p ^ ~q) U (~p -> ~q ) in the case FT (i.e., p false, q true).

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Your solution:

false and true or true then false

confidence rating #$&*:2

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Given Solution:

To evaluate the expression we must first evaluate p ^ ~q and ~p -> ~q.

p ^ ~q is evaluated by first determining the values of p and ~q. If p is false and q true, then ~q is false. Thus both p and ~q are false, and p ^ ~q is false.

~p -> ~q will be false if ~p is true and ~q is false; otherwise it will be true. In the FT case p is false to ~p is true, and q is true so ~q is false. Thus it is indeed the case the ~p -> ~q is false.

(p ^ ~q) U (~p -> ~q ) will be false if (p ^ ~q) and (~p -> ~q ) are both false, and will otherwise be true. In the case of the FT truth values we have seen that both (p ^ ~q) and (~p -> ~q ) are false, so that (p ^ ~q) U (~p -> ~q ) is false.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).

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Your solution:

p q ~p ~q (p^~q) (~p->~q) (p^~q)U(~p->~q)

T T F F F T T

T F F T T T T

F F T T F T T

F T T F T F F

confidence rating #$&*:2

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Given Solution:

We will need headings for p, q, ~p, ~q, (p ^ ~q), (~p -> ~q ) and (p ^ ~q) U (~p -> ~q ). So we set up our truth table

p q ~p ~q (p ^ ~q) (~p -> ~q ) (p ^ ~q) U (~p -> ~q )

T T F F F T T

T F F T T T T

F T T F F F F

F F T T F T T

To see the first line, where p and q are both T, we first see that ~p and ~q must both be false. (p ^ ~q) will therefore be false, since ~q is false; (~p -> ~q) is of the form F -> F and is therefore true. Since (~p -> ~q) is true, (p ^ ~q) U (~p -> ~q ) must be true.

To see the second line, where p is T and q is F, we for see that ~p will be F and ~q true. (p ^ ~q) will therefore be true, since both p and ~q are true; (~p -> ~q) is of the form F -> T and is therefore true. Since (p ^ ~q) and (~p -> ~q ) are both true, (p ^ ~q) U (~p -> ~q ) is certainly true.

To see the fourth line, where p is F and q is F, we for see that ~p will be T and ~q true. (p ^ ~q) will be false, since p is false; (~p -> ~q) is of the form T -> T and is therefore true. Since (~p -> ~q ) is true, (p ^ ~q) U (~p -> ~q ) is true.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q005. If we have a compound sentence consisting of three statements, e.g., p, q and r, then what possible combinations of truth values can occur?

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Your solution:

pq pr qr

TT TT TT

FF FF FF

TF TF TF

FT FT FT

confidence rating #$&*:2

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Given Solution:

A compound statement with two statements p and q has four possible combinations of truth values: TT, TF, FT, FF. Here we also have r, which can be either T or F. So we can append either T or F to each of the possible combinations for p and q.

If r is true then we have possible combinations TT T, TF T, FT T, FF T. If r is false we have TT F, TF F, FT F, FF F. This gives us 8 possible combinations: TTT, TFT, FTT, FFT, TTF, TFF, FTF, FFF.

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Self-critique (if necessary):

I got 12 and not sure why not

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Self-critique Rating:

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Question: `q006. Evaluate the TFT, FFT and FTF lines of the truth table for (p ^ ~q) -> r.

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Your solution:

p q r ~q (p^~Q)(p^~q)->r

T F T T T T

F F T T F T

F T F F F T

confidence rating #$&*:2

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Given Solution:

We would need column headings p, q, r, ~q, (p^~q) and (p^~q) -> r. The truth table would then read

p q r ~q (p^~q) (p^~q) -> r

T F T T T T

F F T T F T

F T F F F T

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q007. Construct a truth table for the statement ~p -> q

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Your solution:

p q ~p ~p->q

T T F F

F F T T

T F T T

F T T T

@&

In your first line ~p is false. A conditional is false only in the case where the antecedent is true and the consequent is false.

Since the antecedent ~p is false, it follows that the conditional cannot be false. It is therefore true in the case of the first line.

There is a line in which the conditional is false but it is not the first.

*@

confidence rating #$&*:

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Self-critique Rating:

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&#This looks good. See my notes. Let me know if you have any questions. &#