34 qa 1

#$&*

course Mth 151

843 11/7/12

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

.

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 6 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

T F F

F T T

F F T

confidence rating #$&*:

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3

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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):

ok

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Self-critique Rating:

3

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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

T F F T

F T T T

F F T F

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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):

ok

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Self-critique Rating:

3

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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:

(f ^ ~) V (~f ->~t)

(F ^ F) V (t -> f)

F V F

False

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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):

ok

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Self-critique Rating:

3

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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 ) V (~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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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):

ok

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Self-critique Rating:

3

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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:

2^3= 8

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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):

Ok, but I didn’t list all of them, would that be a requirement on a test?

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Self-critique Rating:

@&

You could be asked to list them all, but this question just asked for the number, so your answer would be OK.

*@

3

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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 ~q (P ^ ~ Q) -> r

T T F F F

T F T T T

F T F F T

F F T F T

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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):

I don’t know where the R comes in or how to tell if it’s true or false. I am sure I am missing some part of this???

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Self-critique Rating:0

@&

In the preceding exercise you indicated that there would be 8 possible combinations of the headings.

You should know what those 8 combinations would be. Listing them in order p, q, r they would be

TTT

TTF

TFT

TFF

FTT

FTF

FFT

FFF

A complete truth table for any statement involving p, q and r would include all these values, followed in order by the values of ~q, p ^ ~q and (p ^ ~q) -> r.

On this question you were asked only to evaluate the expression for the lines TFT, FFT and FTF. The table in the given solution shows these three lines.

*@

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#