Assignment 15

course Mth 151

KwfQȠ~assignment #015

015. Conditionals

Liberal Arts Mathematics I

05-15-2009

......!!!!!!!!...................................

10:24:41

`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.

......!!!!!!!!...................................

RESPONSE -->

P Q P -> Q

T T T

T F F

F T T

F F T

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:24:47

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

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:27:05

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

......!!!!!!!!...................................

RESPONSE -->

P Q P->Q ~P ~P->Q

T T T F T

T F F F T

F T T T T

F F T T F

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:27:21

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

......!!!!!!!!...................................

RESPONSE -->

mmmmm I'm not sure why

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:28:41

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

......!!!!!!!!...................................

RESPONSE -->

T

confidence assessment: 2

.................................................

......!!!!!!!!...................................

10:29:36

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.

......!!!!!!!!...................................

RESPONSE -->

I thought that the first part would be false and the second part would be true because I didn't see the ~ before the p of the second part

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:31:17

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

......!!!!!!!!...................................

RESPONSE -->

P Q (P ^ ~Q) U ( ~P -> ~Q)

T F T

T T T

F T T

F F T

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:31:59

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 F 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.

......!!!!!!!!...................................

RESPONSE -->

I condensed it

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:32:47

`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?

......!!!!!!!!...................................

RESPONSE -->

TTT

TTF

TFT

FTT

TFF

FFT

FTF

FFF

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:32:55

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.

......!!!!!!!!...................................

RESPONSE -->

got it :-)

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:33:43

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

......!!!!!!!!...................................

RESPONSE -->

T

T

T

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:34:00

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

......!!!!!!!!...................................

RESPONSE -->

I don't know what the answer is. It doesn't say.

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:34:09

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

......!!!!!!!!...................................

RESPONSE -->

Never mind haha

self critique assessment: 3

.................................................

&#Your work looks very good. Let me know if you have any questions. &#