PHIL 4160: Symbolic Logic Dr. Robert Lane Lecture Notes: Tuesday February 21, 2012

[5.3.] “Strategy Hints for Using CP and IP.” (Ch.5:3)

This short section of chapter 5 (5:3, p.139) provides the following hints:

Conditional Proof

§  If your conclusion is a conditional, try using CP.

§  Try using CP if your conclusion is not a conditional but if it equivalent to a conditional; e.g., if your conclusion has the form ~p Ú q, you can use CP to prove p É q, then use Impl to convert that conditional to the conclusion you want.

§  Sometimes it makes for a shorter proof to assume the negation of the consequent of the conditional you want to prove (i.e., assume ~q), derive the negation of the antecedent (~p) thereby proving ~q É ~p, and then use Contra to convert that conditional to p É q.

Indirect Proof

§  “If all else fails, try using IP” -- this is because all proofs can be solved using this rule (although not all proofs can be solved in the fewest steps using it).

§  IP is the preferred method of solving a proof when the conclusion is an atomic sentence or the negation of an atomic sentence.

Exercise 5-5 (pp.137-138)

§  complete all of these; we’ll check some of the odds next time

[5.4.] “Zero-Premise Deductions.” (Ch.5:4)

In this section, you will learn a new use for the proof method:  to prove that a sentence is a theorem of sentential logic.

theorem: “a tautology derived by a zero-premise deduction” (150)

Up to now, we have not encountered a zero-premise deduction. All of the proofs you’ve worked up to this point have had at least one undischarged premise, i.e., at least one premise that is given at the beginning of the proof itself and that remains, undischarged, after any CPs or IPs are completed.

Recall that a premise is discharged when it is set aside as no longer being eligible for use in the proof. Anytime you use CP or IP, you begin by assuming a premise. That premise and all premises that follow from it must be discharged before the proof is complete. In all the proofs we’ve examined up to now, there has always remained behind at least one premise with which you started and which remained undischarged.

If you can prove a conclusion using CP or IP, based on nothing but premises that you yourself assume (i.e., without any starting premises), then you will have shown that sentence to be a theorem of sentential logic.

In exercise 5-6 (p.139) you are asked to prove that a given sentence is a theorem.

·         You do this by constructing a proof that has no starting premises.

·         To begin the proof, just use CP or IP as you normally would.

For example:

Problem #2 asks you to show that (A · B) É A is a theorem. You do this by beginning a proof with an assumed premise, just as in past uses of CP:

(2)        1. A · B                       AP / \ A

2. A                             1 Simp

3. (A · B) É A             1-2 CP

Problem #4 asks you to show that [A É (B É C)         ] É [(A É B) É (A É C)] is a theorem. You do this by beginning a proof with an assumed premise:

(4)        1. A É (B É C)                                                            AP / \ (A É B) É (A É C)]

2. A É B                                                          AP /\ A É C

3. A                                                                 AP / \ C

4. B É C                                                          1, 3 MP

5. A É C                                                          2, 4 HS

6. C                                                                  3, 5 MP

7. A É C                                                          3-6 CP

8. (A É B) É (A É C)                                      2-7 CP

9. [A É (B É C)           ] É [(A É B) É (A É C)]          1-8 CP

Exercise 5-6 (p.139):

·         do all of these; we’ll check some of the odds next time

[5.5.] “Proving Premises Inconsistent.”

Any set of sentences is either consistent (it is possible for all the sentences to be true at the same time) or inconsistent (it is impossible for all the sentences to be true at the same time).

Now we will use the proof method to demonstrate that a set of premises is inconsistent. We will do this by deriving a contradiction from those premises.

Remember that a contradiction is a sentence that cannot possibly be true. So if a set of sentences implies a contradiction, then at least one of those sentences must be false. So to demonstrate that a set of sentences implies a contradiction is to show that it is impossible for all of those sentences to be true at the same time. It may still be possible for some of them to be true, but it is impossible for them all to be true simultaneously.

For example, let’s do #2 in exercise 5-7 (p.140). You are given an argument:

1.      ~A Ú B                   p

2.      ~B Ú ~A                 p

3.      A                           p /\ B

and asked to show that the three premises are inconsistent. To do this, use the rules that you’ve learned to derive a contradiction (a sentence of the form p · ~p) from the premises (important: the conclusion of the argument is irrelevant; to prove the premises to be inconsistent, you need only derive a contradiction from the premises themselves, so in these problems you should simply ignore the conclusion of the argument you are given):

4. ~~A                         3 DN

5. ~B                            2, 4 DS

6. ~A                           1, 5 DS

7. A · ~A                     3, 6 Conj

Exercise 5-7, p.140

§  do all of these for next time; we’ll check some of the odd problems in class

[5.6.] Strange Types of Valid Argument. [Chapter 3:7]

In part B of exercise 5-7, the textbook asks this question about the arguments given in part A: “You can know that all these arguments are valid without proving any of them. Why? How?” (p.140)

To see the answer to this question, we need to go back to chapter three and work through some material we skipped the first time around…

In ch.3:7 (beginning on p.74), your textbook reminds you that a valid argument is one for which there is no assignment of truth values to constants on which all the premises are true and the conclusion false at the same time.

A complete truth table for the valid argument   p, ~p Ú q / \ q   demonstrates this:

 p q p ~p Ú q q T T T FT T T T T F T FT F F F F T F TF T T T F F F TF T F F

There is no row (and so no assignment of truth values) on which the premises are true and the conclusion false. So the argument is valid. It is important to note that it is the absence of a certain kind of row that shows that the argument is valid, namely, the absence of a row with all true premises and a false conclusion.

This definition of validity implies that there are three types of valid argument:

A.     “the normal case”: all three of the following claims are true about a “normal” valid argument:

1)      it is possible for the premises to be all true (the premises are consistent);

2)      it is possible for the conclusion to be false (the conclusion is not a tautology); and

3)      it is not possible for the premises to be all true and the conclusion false at the same time.

B.     the first “special case”: it is impossible for the premises to be all true, i.e., the premises are inconsistent. In this case, there is no assignment of truth values on which the premises are all true… and therefore no assignment of truth values on which the premises are all true and the conclusion false.

C.     the second “special case”: it is impossible for the conclusion to be false, i.e., the conclusion is a tautology. In this case, there is no assignment of truth values on which the conclusion is false… and therefore no assignment of truth values on which the premises are all true and the conclusion false.

So any argument with inconsistent premises is valid; and any argument that has a tautology as its conclusion is valid. Thus, the answer to exercise 5-7 sec.B is: you can know that all of the arguments in sec.A are valid because each of them has inconsistent premises. Each argument is an example of the first kind of “special case” valid argument.

[5.7.] The Short Truth Table Test for Invalidity. [Ch. 3:8]

The proof method is an extremely efficient way to demonstrate that an argument is valid. But you cannot use it to demonstrate that an argument is invalid. To show an argument to be invalid, we still need to rely on the truth table method.

But we don’t have to use the full truth table method covered a few weeks ago. Instead, we can use the more economical short truth table method.

I will explain this method in the same way your textbook does, by applying it to a specific example: exercise 3-10 (p.79),  #4:

(4)         1. R Ú N

2. L É N

3. R                  /\ ~N

Step 1: Set up the top row of a truth table, containing just the premises and conclusion:

 R Ú N L É N R ~N

(Notice that, unlike in the complete truth table method, the top row of the table consists of the premises themselves rather than their one-one logical forms—it thus contains sentence constants rather than sentence variables.)

Step 2: Assign truth values to the main connectives of each sentence, or if a sentence consists of a single constant, assign a truth value to that constant. Assign the value “true” to each premise and assign the value “false” to each conclusion.

 R Ú N L É N R ~N T T T F

Step 3: if you have assigned a value to any sentence constants, place those values elsewhere on the table where that constant occurs.

 R Ú N L É N R ~N T T _ T T F

Step 4: Now begin filling in the spaces underneath the other operators and/or constants, according to the truth table definitions of the constants.

In this example, you should first insert the value of the “N” in the conclusion, since there is only one value it can take:

 R Ú N L É N R ~N T T _ T T FT

Then add “T” underneath every other occurrence of “N” in the proof:

 R Ú N L É N R ~N T T T _ T T T FT

And since a conditional with a true consequent is true whether its antecedent is true or false, you can fill in the value of “L” in the second premise either way; here I will fill in “F”, although “T” would work just as well:

 R Ú N L É N R ~N T T T F T T T FT

Once you’ve completed the short truth table in this manner, you have shown that it is possible for all the premises of the argument to be true and the conclusion false at the same time. In other words, you’ve shown that the argument is invalid.

IMPORTANT NOTE: Sometimes you will have a choice as to whether to assign a T or an F to a given constant (as in the assignment of “F” to “L” in the above example). This is because, with regard to some compound sentences, you can change the truth value of a component sentence without changing the truth value of the entire sentence. For example, suppose that “A” is true. In that case, “A Ú B” is true no matter what the truth value of “B” happens to be. In such cases, at least one of the truth values will allow you to complete the truth table. Choose one, and if it doesn’t work, proceed to try the next one. (See the examples on pp.78 of your textbook.)

Exercise 3-10 (p.79)

·         [we will do one or two evens in class]

·         do all of these for next time; we’ll check some of the odds in class

·         make sure you write out the entire short truth table (unlike in the back of the book)

[5.8.] The Short Truth Table Test for Consistency. [Ch. 3:9]

We can also use the short truth table method to demonstrate that the premises of an argument are consistent, i.e., that it is possible for them all to be true at the same time. Use the same steps as those described in the last section, except don’t bother including the conclusion in the top row of your table.

For example: exercise 3-11 (p.82), #2:

(2)               1. ~(F º G)

2. ~(F É H)                  /\ F

Step 1:

 ~(F º G) ~(F É H)

Step 2:

 ~(F º G) ~(F É H) T _ _ _ T _ _ _

Step 3: [no values have been assigned to constants at this point]

Step 4:

 ~(F º G) ~(F É H) T _ F _ T _ F _

 ~(F º G) ~(F É H) T _ F _ T T F F

 ~(F º G) ~(F É H) T T F F T T F F

Notice that on this last step, when you assign value T to constant “F” and value F to constant “G”, this is consistent with the truth table definition of the tri-bar. You could not assign the same truth value to both “F” and “G”, since that would make the biconditional true, rending the entire premise false.

Exercise 3-11 (p.82):

·         complete all of these for next time

·         we will cover some of the odds in the next class

·         make sure you write out the entire short truth table (unlike in the back of the book)

Stopping point for Tuesday February 21. For next time:

·         complete exercises 5-5, 5-6, 5-7, 3-10, and 3-11.

·         no new reading; your second test is one week from today; study guide is online.

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