assignment 11 compound sentences algebra 2

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2.1.5: Compounding Compound Sentences

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We have seen how to apply the connectives '~', '&', and 'v' to atomic sentences such as 'A' and 'B' to get compound sentences such as '~A', 'A&B', and 'AvB'. But could we now do this over again? That is, could we apply the connectives not just to atomic sentences 'A', 'B', 'C', etc., but also to the compound sentences '~A', 'A&B', and 'AvB'? Yes, of course. For example, we can form the conjunction of '~A' with 'B', giving us '~A&B'. Using our current transcription guide, this transcribes into 'Adam does not love Eve and Adam is blond.'

As another example, we could start with the conjunction 'A&B' and take this sentence's negation. But now we have a problem. (This is the problem you encountered in trying to work exercise 1~2, e~i.) If we try to write the negation of 'A&B' by putting a '~' in front of 'A&B', we get the sentence we had before. But the two sentences should not be the same! This might be a little confusing at first. Here is the problem: We are considering two ways of building up a complex sentence from shorter parts, resulting in two different complex sentences. In the first way, we take the negation of 'A', that is, '~A', and conjoin this with 'B'. In the second way, we first conjoin 'A' and 'B' and then negate the whole. In English, the sentence 'It is not the case both that Adam loves Eve and Adam is blond.' is very different from the sentence 'Adam does not love Eve, and Adam is blond.' (Can you prove this by giving circumstances in which one of these compound sentences is true and the other one is false?)

In order to solve this problem, we need some device in logic which does the work that 'both' does in English. (If you are not sure you yet understand what the problem is, read the solution I am about to give and then reread the last paragraph.) What we need to do is to make clear the order in which the connectives are applied. It makes a difference whether we first make a negation and then form a conjunction, or whether we first form the conjunction and then make a negation. We will indicate the order of operations by using parentheses, much as one does in algebra. Whenever we form a compound sentence we will surround it by parentheses. Then you will know that the connective inside the parentheses applies before the one outside the parentheses. Thus, when we form the negation of 'A', we write the final result as '(~A)'. We now take '(~A)' and conjoin it with 'B', surrounding the final result with parentheses:

\[ [(~A}\&B] \label{4}\]

This says, take the sentence '(~A)' and conjoin it with 'B'. To indicate that the final result is a complete sentence (in case we will use it in some still larger compound), we surround the final result in parentheses also. Note how I have used a second style for the second pair of parentheses-square brackets-to make things easier to read.

Contrast \ref{4}4)

\[[(A\&B)] \label{5}\]

with which means that one is to conjoin 'A' with 'B' and then take the negation of the whole.

In the same kind of way we can compound disjunctions with conjunctions and conjunctions with disjunctions. For example, consider

\[[(A\&B)vC] \label{6}\]

\[ \label{7} [(A\&(BvC))]\]

Sentence (6) says that we are first to form the conjunctions of 'A' with 'B' and then form the disjunction with 'C'. Expression \ref{7}, on the other hand, says that we are first to form the disjunction of 'B' with 'C' and then conjoin the whole with 'A'. These are very different sentences. Transcribed into English, they are 'Adam both loves Eve and is blond, or Eve is clever.' and 'Adam loves Eve, and either Adam is blond or Eve is clever.'

We can show more clearly that Expression \ref{6} and \ref{7} are different sentences by writing out truth tables for them. We now have three atomic sentences, 'A', 'B', and 'C'. Each can be true or false, whatever the others are, so that we now have eight possible cases. For each case we work out the truth value of a larger compound from the truth value of the parts, using the truth value of the intermediate compound when figuring the truth value of a compound of a compound:

Let's go over how we got this truth table. Columns a, b, and c simply give all possible truth value assignments to the three sentence letters 'A', 'B', and 'C'. As before, in principle, the order of the cases does not matter. But to make it easy to compare answers, you should always list the eight possible cases for three letters in the order I have just used. Then, for each case, we need to calculate the truth value of the compounds in columns d through h from the truth values given in columns a, b, and c.

Let us see how this works for case 5, for example. We first need to determine the truth value to put in column d, for '(A&B)' from the truth values given for this case. In case 5 'A' is false and 'B' is true. From the truth table definition of '&', we know that a conjunction (here, 'A&B') is false when the first conjunct (here, 'A') is false and the second conjunct (here, 'B') is true. So we write an 'f' for case 5 in column d. Column e is the disjunction of 'B' with 'C'. In case 5 'B' is true and 'C' is true. When we disjoin something true with something true, we get a true sentence. So we write the letter 't', standing for the truth value t, in column e for case 5.

Moving on to column g, we are looking at the disjunction of '(A&B)' with 'C'. We have already calculated the truth value of '(A&B)' for case 5~that was column d~and the truth value of 'C' for case 5 is given in column c. Reading off columns c and d, we see that '(A&B)' is false and 'C' is true in case 5. The sentence of column g, '[(A&B)vC]', is the disjunction of these two components and we know that the disjunction of something false with something true is, again, true. So we put a 't' in column g for case 5. Following the same procedure for column h, we see that for case 5 we have a conjunction of something false with something true, which gives the truth value f. So we write 'f for case 5 in column h.

Go through all eight cases and check that you understand how to determine the truth values for columns d through h on the basis of what you are given in columns a, b, and c.

Now, back to the point that got us started on this example. I wanted to prove that the sentences '[(A&B)vC]' and '[A&(BvC)]' are importantly different. Note that in cases 5 and 7 they have different truth values. That is, there are two assignments of truth values to the components for which one of these sentences is true and the other is false. So we had better not confuse these two sentences. You see, we really do need the parentheses to distinguish between them.

Actually, we don't need all the parentheses I have been using. We can make two conventions which will eliminate the need for some of the parentheses without any danger of confusing different sentences. First, we can eliminate the outermost parentheses, as long as we put them back in if we decide to use a sentence as a component in a larger sentence. For example, we can write 'A&B' instead of '(A&B)' as long as we put the parentheses back around 'A&B' before taking the negation of the whole to form '~(A&B)'. Second, we can agree to understand '~' always to apply to the shortest full sentence which follows it. This eliminates the need to surround a negated sentence with parentheses before using it in a larger sentence. For example, we will write '~A&B' instead of '(~A)&B'. We know that '~A&B' means '(~A)&B' and not '~(A&B)' because the '~' in '~A&B' applies to the shortest full sentence which follows it, which is 'A' and not 'A&B'.

This section still needs to clarify one more aspect of dealing with compound sentences. Suppose that, before you saw the last truth table, I had handed you the sentence '(A&B)vC' and asked you to figure out its truth value in each line of a truth table. How would you know what parts to look at? Here's the way to think about this problem. For some line of a truth table (think of line 5, for example), you want to know the truth value of '(A&B)vC'. You could do this if you knew the truth values of 'A&B' and of 'C'. With their truth values you could apply the truth table definition of 'v' to get the truth value of '(A&B)vC'. This is because '(A&B)vC' just is the disjunction of 'A&B' with 'C'. Thus you know that '(A&B)vC' is true if at least one of its disjuncts, that is, either 'A&B' or 'C', is true; and '(A&B)vC' is false only if both its disjuncts, 'A&B' and 'C', are false.

And how are you supposed to know the truth values of 'A&B' and of 'C'? Since you are figuring out truth values of sentences in the line of a truth table, all you need do to figure out the truth value of 'C' on that line is to look it up under the 'C' column. Thus, if we are working line 5, we look under the 'C' column for line 5 and read that in this case 'C' has the truth value t. Figuring out the truth value for 'A&B' for this line is almost as easy. 'A&B' is, by the truth table definition of conjunction, true just in case both conjuncts (here, 'A' and 'B') are true. In line 5 'A' is false and 'B' is true. So for this line, 'A&B' is false.

Now that we finally have the truth values for the parts of '(A&B)vC', that is, for 'A&B' and for 'C', we can plug these truth values into the truth table definition for v and get the truth value t for '(A&B)vC'.

Now suppose that you have to do the same thing for a more complicated sentence, say

\[ \label{8} ~{[Av~C\]&[Bv(~A&C)]} \]

Don't panic. The principle is the same as for the last, simpler example. You can determine the truth value of the whole if you know the truth value of the parts. And you can determine the truth value of the parts if you can determine the truth value of their parts. You continue this way until you get down to atomic sentence letters. The truth value of the atomic sentence letters will be given to you by the line of the truth table. With them you can start working your way back up.

You can get a better grip on this process with the idea of the Main Connective of a sentence. Look at sentence (8) and ask yourself, 'What is the last step I must take in building this sentence up from its parts?" In the case of (8) the last step consists in taking the sentence '[Av~C]&[Bv(~A&C)]' and applying '~' to it. Thus (8) is a negation, '~' is the main connective of (a), and '[Av~C]&[Bv(~A&C)]' is the component used in forming (8).

What, in turn, is the main connective of '[Av~C]&[Bv(~A&C)]'? Again, what is the last step you must take in building this sentence up from its parts? In this case you must take 'Av~C' and conjoin it with 'Bv(~A&C)'. Thus this sentence is a conjunction, '&' is its main connective, and its components are the two conjuncts 'Av~C' and 'Bv(~A&C)'. In like manner, 'Bv(~A&C)' is a disjunction, with 'v' its main connective, and its components are the disjuncts 'B' and '~A&C'. To summarize,

The Main Connective in a compound sentence is the connective which was used last in building up the sentence from its component or components.

Now, when you need to evaluate the truth value of a complex sentence, given truth values for the atomic sentence letters, you know how to proceed. Analyze the complete sentence into its components by identifying main connectives. Write out the components, in order of increasing complexity, so that you can see plainly how the larger sentences are built up from the parts. In the case of (8), we would lay out the parts like this:

A, B, C, ~A, ~C, Av~C, ~A&C, Bv(~A&C), [Av~C]&[Bv(~A&C)], ~[Av~C]&[Bv(~A&C)]}

You will be given the truth values of the atomic sentence letters, either by me in the problem which I set for you or simply by the line of the truth table which you are working. Starting with the truth values of the atomic sentence letters, apply the truth table definitions of the connectives to evaluate the truth tables of the successively larger parts.

Exercise \(\PageIndex{1}\)

For each of the following sentences, state whether its main connective is '~', '&', or 'v' and list each sentence's components. Then do the same for the components you have listed until you get down to atomic sentence letters. So you can see how you should present your answers, I have done the first problem for you.

• (D&~G)v(G&D)
• [(Dv~~B)&(DvB)]&(DvB)
• L&[Mv[~N&(Mv~L)}}

Contributors and Attributions

Paul Teller  (UC Davis). The Primer was  published  in 1989 by Prentice Hall, since acquired by Pearson Education. Pearson Education has allowed the Primer to go out of print and returned the copyright to Professor Teller who is happy to make it available without charge for instructional and educational use.

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1.1: Compound Statements

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• Pamini Thangarajah
• Mount Royal University

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We can make a new statement from other statements; we call these compound propositions or compound statements .

Example \(\PageIndex{1}\):

• It is not the case that all birds can fly. (This is the negation of the statement all birds can fly).
• \(1+1=2\) and "All birds can fly". (Here the connector "and" was used to create a new statement).

Note the following four basic ways to start with one or more propositions and use them to make a more elaborate compound statement. If \(p\) and \(q\) are statements then here are four compound statements made from them:

• \(\neg p \), Not \(p\) (i.e. the negation of \(p\)),
• \( p \wedge q,\, p\, \textit{and}\, q\),
• \(p\vee q, \,p \,\textit{or} \,q\) and
• \(p \rightarrow q,\: \textit{If} \; p \, \textit{then}\, q.\)

Example \(\PageIndex{2}\):

If \(p =\) "You eat your supper tonight" and \(q = \) "You get desert". Then

• Not \(p \) is "You don't eat your supper tonight".
• \(p\, \textit{and}\, q\) is "You eat your supper tonight and you get desert".
• \( p \,\textit{or} \,q\) is "You eat your supper tonight or you get desert".
• \(\textit{If} \; p \, \textit{then}\, q\) is "If you eat your supper tonight then you get dessert."

In English, we know these four propositions don't say the same thing. In logic, this is also the case, but we can make that clear by displaying the truth value possibilities. It is common to use a table to capture the possibilities for truth values of compound statements. We call such a table a truth table. Below are the possibilities: the first is the least profound. It says that a statement p is either true or false.

Truth tables are more useful in describing the possible truth values for various compound propositions. Consider the following truth table:

The table above describes the truth value possibilities for the statements \(p\) and \(\neg p\), or "not p". As you can see, if \(p\) is true then \(\neg p\) is false and if \(p\) is false, the negation (i.e. not p) is true. \(\neg\) is the mathematical notation used to mean "not."

Example \(\PageIndex{3}\):

Consider the statement \(p\): \(1 + 1 = 3\).

Statement \(p\) can either be true or false, not both.

\(\neg p\) is "not \(p\)," or the negation of statement \(p\).

\(\neg p\) is \(1 + 1 \ne 3\).

You can see that the negation of a proposition affects only the proposition itself, not any other assumptions.

Conjunction

Conjunction statements use two or more propositions. If two or more simple propositions are involved the truth table gets bigger. Below is the truth table for "and," otherwise known as a conjunction. When is an and statement true? As the truth table indicates, only when both of the component propositions are true is the compound conjunction statement true:

Example \(\PageIndex{4}\):

Consider statements \(p:= \,1 + 1 = 2\) and \(q:=\,2 < 5\).

Note that, \(p \wedge q\) is true only if both \(p\) and \(q\) are both true.

Since statements \(p\) and \(q\) are both true, \(p \wedge q\) is true.

Disjunction

Disjunction statements are compound statements made up of two or more statements and are true when one of the component propositions is true. They are called "Or Statements." In English, "or" is used in two ways:

• If a person is looking for a house with 4 bedrooms or a short commute, a real estate agent might present houses with either 4 bedrooms or a short commute or both 4 bedrooms and a short commute. This is called an inclusive or .
• If a person is asked whether they would like a Coke or a Pepsi, they are expected to choose between the two options. This is an exclusive or : "both" is not an acceptable case.

In logic, we use inclusive or statements

The \(p \) or \( q\) proposition is only false if both component propositions \(p \) and \( q\) are false.

Example \(\PageIndex{5}\):

Consider the statement \(2 \leq -3\)

The statement reads "2 is less than or equal to -3", or "\(2 < -3 \vee 2 = -3\)" and can be broken into two component propositions:

• Proposition \(p\): \(2 < -3\) (False)
• Proposition \(q\): \(2 = -3\) (False)

Because propositions \(p\) and \(q\) are both false, the statement is false.

Example \(\PageIndex{6}\):

Consider the statement \(2 \leq 5\)

The statement's two component propositions are:

• Proposition \(p\): \(2 < 5\) (True)
• Proposition \(q\): \(2 = 5\) (False)

Since proposition \(p\) is true, the statement is true.

Conditional Statements

Consider the "if p then q" proposition. This is a conditional statement . Read the statements below. If these statements are made, in which instance is one lying (i.e. when is the overall statement false)?

Suppose, at suppertime, your mother makes the statement “If you eat your broccoli then you’ll get dessert.” Under what conditions could you say your mother is lying?

• If you eat your broccoli but don't get dessert, she lied!
• If you eat your broccoli and get dessert, she told the truth.
• If you don’t eat your broccoli and you don’t get dessert she told you the truth.
• If you don’t eat your broccoli but you do get dessert we still think she told the truth. After all, she only outlined one condition that was supposed to get you desert, she didn’t say that was the only way you could earn dessert. Maybe you had cauliflower instead.

Note that the order in which the cases are presented in the truth table is irrelevant. The cases themselves are important information, not their order relative to each other.

It is important to notice that, if the first proposition is false, the conditional statement is true by default. A conditional statement is defined as being true unless a true hypothesis leads to a false conclusion.

Example \(\PageIndex{7}\):

Consider the statement "If a closed figure has four sides, then it is a square." This is a false statement - why?

We can prove it using a counter-example : we draw a four-sided figure that is not a square. So there!

Example \(\PageIndex{8}\):

Consider the statement "If \(2 = 3\), then \(5 = 2\)"

Since \(2 \ne 3\), it does not matter if \(5 = 2\) is true or not, the conditional statement as a whole is true.

The converse of a conditional statement

Let P be a statement if p then q. Then the converse of P is if q then p.

Example \(\PageIndex{9}\):

Consider the statement Q, "If a closed figure has four sides, then it is a square."

Then the converse of Q is "If it is a square then it is a closed figure with four sides".

The contrapositive of a Conditional Statement

Let P be a statement if p then q. Then the contrapositive of P is if \(\neg q\) then \(\neg p.\)

Example \(\PageIndex{10}\):

Then the converse of Q is "If it is not a square then it is not a closed figure with four sides".

Bi-Conditional Statements

Bi-conditional statements are conditional statements which depend on both component propositions. They read "p if and only if q" and are denoted \(p \leftrightarrow q\) or "p iff q", which is logically equivalent to \((p \to q) \wedge (q \to p)\). These compound statements are true if both component propositions are true or both are false:

Example \(\PageIndex{11}\):

Consider the statement: "Two lines are perpendicular if and only if they intersect to form a right angle."

The component propositions are:

• \(p\): Two lines are perpendicular
• \(q\): [The lines] intersect to form a right angle

Logically, we can see that if two lines are perpendicular, then they must intersect to form a right angle. Also, we can see that if two lines form a right angle, then they are perpendicular.

If two lines are not perpendicular, then they cannot form a right angle. Conversely, if two lines do not form a right angle, they cannot be perpendicular. This is why, if both propositions in a biconditional statement are false, the statement itself is true!

Logically Equivalent Statements

Once we know the basic statement types and their truth tables, we can derive the truth tables of more elaborate compound statements. Below is the truth table for the proposition, not p or (p and q) . First, we calculate the truth values for not p, then p and q and finally, we use these two columns of truth values to figure out the truth values for not p or (p and q).

So the proposition "not p or (p and q)" is only false if p is true and q is false. Does this seem familiar?

"If p then q" is only false if p is true and q is false as well.

This has some significance in logic because if two propositions have the same truth table they are in a logical sense equal to each other – and we say that they are logically equivalent. So: \(\neg p \vee (p \wedge q) \equiv p \to q\), or "Not p or (p and q) is equivalent to if p then q."

Example \(\PageIndex{12}\):

Prove or disprove: for any mathematical statements \(p,q\) and \(r,\, p\to(q \vee r)\) is logically equivalent to \(\neg r \to ( p \to q).\)

Hence, \(p\to(q \vee r)\) is logically equivalent to \(\neg r \to ( p \to q).\)

Tautologies and Contradictions

There are two cases in which compound statements can be made that result in either always true or always false. These are called tautologies and contradictions , respectively. Let's consider a tautology first, and then a contradiction:

Example \(\PageIndex{13}\):

Consider the statement "\((2 = 3) \vee (2 \ne 3)\)":

There are two component propositions:

• \(p\): \(2 = 3\)
• \(\neg p\): \(2 \ne 3\)

Clearly, this statement is a tautology.

Let's make a truth table for general case \(p \vee (\neg p)\):

As you can see, no matter what we do, this statement is always true. It is a tautology . Careful! This is not to say that this statement makes logical sense in English, but rather that, using logical mathematics, this statement is always true.

Example \(\PageIndex{14}\):

Consider the statement "2 is even \(\wedge\) 2 is odd"

• \(p\): 2 is even
• \(\neg p\): 2 is odd

Clearly, this statement is a contradiction.

Let's make a truth table for general case \(p \wedge (\neg p)\):

As you can see again, no matter what we do, this statement will always be false. It is a contradiction . These make more sense in English: 2 cannot be both even and odd, after all! Still, what matters is what we decide using logical mathematics.

Notations & Definitions:

• Negation: \(\neg\) or " not "
• Conjunction: \(\wedge\) or " and "
• Disjunction: \(\vee\) or " or "
• Conditional: \(\to\) or " implies " or " if/then "
• Bi-Conditional: \(\leftrightarrow\) or " if and only if " or " iff "
• Counter-example: An example that disproves a mathematical proposition or statement.
• Logically Equivalent: \(\equiv\) Two propositions that have the same truth table result.
• Tautology: A statement that is always true, and a truth table yields only true results.
• Contradiction: A statement which is always false, and a truth table yields only false results.