Need help with this question - math proofs!

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The P Q ¬P ¬Q P⇒Q P⇔Q P∨Q P∧Q P∧¬Q ¬P∧Q ¬P∧¬Q ¬(P∧¬Q)
T T F F T T T T F F F T
T F F T F F T F T F F F
F T T F T F T F F T F T
F F T T T T F F F F T T

1. From the truth table above, when P and Q are both true, then P⇒Q (if P is true then Q is true), P⇔Q (P is true if and only if Q is true), P∨Q (either or both P or Q are true), so the three logical relationships are equivalent. 

When P∧¬Q (P is true and Q is not true) is true, then we can replace Q in row 1, columns 5, 6, 7 with (P∧¬Q), e.g., P⇔(P∧¬Q), and similarly with the other two logical operations.

What we need to do is to use only ¬ and ∧ with P and Q to find equivalent logical expressions for each of the operations in columns 5, 6, 7. The table should help us to find them.

The last column, as an example, shows what is equivalent to P⇒Q. Here’s how I got the answer.

(I would have added more columns to include the answers I’ve written below, but the truth table would be too wide to read. I hope the explanations below help you to understand the processes.)

P⇒Q gives us the column T F T T. Looking across the table I saw that column P∧¬Q was F T F F which is the complement of what I needed. The NOT operator (¬) converts F to T and T to F, so by writing ¬ in front of P∧¬Q I would get the result I needed. This is written ¬(P∧¬Q).

P⇔Q is a bit trickier, but I got the answer ¬(P∧¬Q)∧¬(¬P∧Q).

P∨Q is ¬(¬P∧¬Q).

2. 

This is what the question is looking for. There are other solutions, so other column  headers are possible. To show you how to do this, let’s take an example. Take column 15, TFTF. In the columns with a header already, we look for another column which already has a header and which looks similar to TFTF. I used the next column 16, P⇒Q: TFTT. All I need is something to change the last T into an F. I also need a logical operation to turn T into F. The operation is AND (∧). T∧F=F. So I noticed that column 19 contained an F and the rest were T’s so combining columns 16 and 19 gave me the result I needed: TFTT ∧ TTTF=TFTF. This gave me (P⇒Q)∧(P∨Q) for column 15’s header. This method turns the problem into a game.

Continued in comment...

 

by Top Rated User (1.1m points)
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So how to do 2& 3?

I’m only part way through the solution. When you see “More to follow...” it means that I haven’t been able (because of editing and network problems) or haven’t yet had time to complete all parts of the question (because I’ve been busy, as an ordinary user, with other things, like shopping, house-keeping, gardening, and so on!).

I recently updated my answer, but that was just question 1, which still needs more to be added. Then I can move on to the other parts. So please be patient! In the meantime, you should be able to see the method I’m using and perhaps continue  working out the solutions for yourself.

[Insufficient space to continue solution.]

3. This time we have three Boolean, P, Q, R so the total number of possible inputs is 2³=8, giving us 8 rows for the table. The possible outputs ranges from FFFFFFFF to TTTTTTTT in the columns, making a total of 2⁸=256 columns. To fill up some of the columns we can P op Q op R where “op” is one of the logical operators, and in addition we have the unary operator NOT (¬) which acts on a single Boolean or the result of a Boolean operation. The other operators are binary, that is, they require two operands. So, to get started, we work out the column outputs for P∨Q∨R, P∨Q∧R, P∨Q⇒R, etc., for combinations of P, Q, R values. So various columns can be completed fairly easily. These are probably the “favourites” referred to in the question. We can also establish whether associativity, commutativity rules can be applied. For example, is (P op Q) op R the same as P op (Q op R), etc.

Commutativity applies to all except implication (⇒); but not associativity in general, so grouping is important.

I would recommend that the favourites include (P∧Q)∨R and P∧(Q∨R) and similar using combinations of different operators because they demonstrate that associativity doesn’t apply generally to logical operators. List all 8 values of (P,Q,R)=(T,T,T) to (F,F,F) and work out the outcomes to give you columns in the truth table. You will become more familiar with logic operations. Please check that I haven’t made any errors by carrying out the logical operations yourself. It’s very easy to get confused!

Here is an example for the answer to question 3:

(The table shows how and where associativity and commutativity do not apply generally to logical operations)

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