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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Self-critique (if necessary):
Self-critique Rating:
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 F
Confidence Assessment:
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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):
I messed up on the last row I thought if p and q were False it would make it False. I know why it would not be false because p is not true.
Self-critique Rating:
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 F
F F T F
Confidence Assessment:
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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):
I am a little confused I am right until row 3 and I am also a little confused about the way the answer is set up like I am missing a step.
There's a typo in my table, which I corrected above (3d row ~p should have been T). With my correction, your table is identical to the given solution except for the last value in the third row. To evaluate ~p -> q in that row you see that ~p is F and q is T. F -> T is T.
Self-critique Rating:
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:
p q ~p ~q p^~q ~p->~q V
T T F F F F F
T F F T T F T
F T T F F T T
F F T T F F T
Confidence Assessment:
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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):
I don? understand the answer. I understand through the table but the way it is written it is confusing. Am I right in my table thinking?
Your table is fine, and it's equivalent to the reasoning in the given solution. You should also try to understand how the given solution is reasoned out, but you'll be OK as long as you understand the tables.
Self-critique Rating:
Question: **** `q004. Construct a truth table for the proposition (p ^ ~q) U (~p -> ~q ).
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Your solution: (I put V instead of U because it makes me think union if I don?)
p q ~p ~q p^~q ~p->~q V
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 Assessment:
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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):
You need to check your table it says the 2nd line p^~q is false. The words it says its true and I also have true.
You're right. Very good. I've made the correction. Thanks.
Self-critique Rating:
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:
p q r ~p ~q ~r
I am really not sure how to go about this
Confidence Assessment:
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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 understand now
Self-critique Rating:
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 ->r
T T F F
T F T T
F T F F
F F T F
I don? know the rule for r
Confidence Assessment:
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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
What is the rule for r so I can finish the table?
You are asked to evaluate the TFT, FFT and TFT lines of the table. So you would start out with
p q r ~q p^~q ->r
T F T
F F T
T F T
and go from there.
&#This looks good. See my notes. Let me know if you have any questions. &#