assn 15

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course Mth 151

Again i found it impossible to show my work on my computer, partly because im not the most computer savy, but i worked all tables out on paper and have a good feeling about it

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:

GOT IT

T

F

T

T

confidence rating #$&* 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):

oik

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Self-critique rating #$&*

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

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Your solution:

T

T

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 #$&*

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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:

FF

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 #$&*

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

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Your solution:

T

T

F

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 #$&*

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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:

TTT

TTF

TFT

TFF

FTT

FTF

FFT

FFF

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

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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:

All are true

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):

ok

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Self-critique rating #$&*

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&#Please let me know if you have questions. &#

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