Assignment 15A

course MTH 151

???x?????M???assignment #015

015. Conditionals

Liberal Arts Mathematics I

03-12-2009

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18:34:36

`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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RESPONSE -->

p q p->q

T T T

T F F

F T T

F F T

confidence assessment: 2

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18:35:12

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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RESPONSE -->

correct

self critique assessment: 3

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18:37:24

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

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RESPONSE -->

p q ~p ~p->q

T T F T TRUE

T F F T TRUE

F T F T TRUE

F F T T FALSE

confidence assessment: 0

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18:37:36

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 F T since (T -> T) is T

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

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RESPONSE -->

correct

self critique assessment: 3

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18:47:13

`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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RESPONSE -->

(p^ ~q) V ( ~ p -> ~q) would be false because

p ^ ~q p is false and ~q is false

~p-> ~q ~p is true ~q if false

confidence assessment: 0

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18:47:30

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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RESPONSE -->

correct

self critique assessment: 3

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18:50:53

`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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RESPONSE -->

I was typing in the answer to this question and pressed a key that took me to the answer screen. I am not sure what happened but I understand the answer.

confidence assessment: 2

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18:51:08

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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RESPONSE -->

ok

self critique assessment: 3

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18:53:20

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

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RESPONSE -->

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 assessment: 1

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18:53:33

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

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RESPONSE -->

correct

self critique assessment: 3

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18:53:41

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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RESPONSE -->

correct

self critique assessment: 3

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&#Very good work. Let me know if you have questions. &#