PHIL 4160: Symbolic Logic Dr. Robert Lane Lecture Notes: Thursday February 16, 2012

[5.] Conditional and Indirect Proofs

[5.1.] “Conditional Proofs.”

[5.1.1.] Introducing CP.

Chapter 5 of your textbook begins with an example of an argument that is valid but that cannot be proved valid using only the rules covered so far (namely, the 8 implicational rules and the 10 equivalence rules):

A É B / \ A É (A · B)

Notice that the conclusion is a conditional: it does not assert that A is true, and it does not assert that A · B is true. What it asserts is that if A is true, then so is A · B.

So in order to derive this conclusion from the premise given, we don’t have to show either that A is true or that A · B is true. We merely have to show that (given the premise A É B) if A is true, then so is A · B.

In more general terms, what we want to prove is that (given specific premises), if one sentence is true, then another is true as well.

Chapter 5:1 introduces a new rule that allows us to prove this sort of conclusion: the rule of Conditional Proof (CP). See the front inside cover of your textbook for the form of this rule.

Here is how the rule works (I will use the argument given above as an example). You begin as with any other proof in sentential logic:

1. A É B                                  p          /\ A É (A · B)

The first step in using CP is to assume the antecedent of the conditional you want to prove (this is the only thing you should ever use as an assumed premise when using CP!):

2. A                                         AP [for “Assumed Premise”] / \ A · B

The second thing to do is derive the consequent of the conditional from the antecedent that you have just assumed. In this case, the consequent is A · B:

3. B                                          1, 2 MP

4. A · B                                   2, 3 Conj

At this point, you have shown that if A is true, then A · B is true as well. In other words, you have shown “A É (A · B)” to be true. This licenses you to write that conditional down in your proof, citing “CP” as the rule you are using to do so:

5. A É (A · B)                         2-4 CP

Note that correct notation requires that you draw a long arrow, beginning above the citation of CP in line five and pointing to the line on which you used AP. See p.124 in your book for an example of this notation.

This notation indicates that the sentence derived by way of CP (in this example, line 5) does not depend on the premise assumed on the line that uses AP (in this example, line 2). In other words, it indicates that line 2 (“A”) does not have to be true in order for line 5 [A É (A · B)] to be true.

It also indicates that everything between line 2 and the horizontal line between lines 4 and 5 does depend on line 2. You could not have proved lines 3 and 4 without assuming line 2, and the arrow indicates this.

Lines 3 and 4 are said to be within the scope of the assumed premise.

Beginning just underneath that horizontal line (from line 5 on), we say that by this point in the proof, the assumed premise has been discharged.

An additional CP proof, which employs some less commonly used rules…

1. (C · D) É B             p / \ D É (~C Ú B)

2. D                             AP / \ ~C V B

3. (D · C) É B             1 Comm

4. D É (C É B)                        3 Exp

5. C É B                      3, 4 MP

6. ~C Ú B                     5 Impl

7. D É (~C Ú B)           2-6 CP

[5.1.2.] Discharged Assumed Premises.

IMPORTANT: Once an assumed premise has been discharged, you can no longer use it, or any of the premises within its scope, in the proof.

For example, you cannot continue the previous proof with the following line:

8. D Ú C                       2 Add

That move employs line 2, which is within the scope of the assumed premise, which at this point in the proof has been discharged.

[5.1.3.] Arguments with Conclusions that are not Conditionals.

CP can be useful, even when the conclusion of your argument is not a conditional. Suppose that you were asked to prove the following argument valid:

A É B / \ ~A Ú (A · B)

You might reason as follows: ~A Ú (A · B) is equivalent to the conditional A É (A · B), and thus can be replaced by that conditional using Impl. So if you can prove A É (A · B), then you are one step away from proving ~A Ú (A · B). Your proof would look like this:

1. A É B                                  p          /\ ~A Ú (A · B)

2. A                                         AP / \ A · B

3. B                                          1, 2 MP

4. A · B                                   2, 3 Conj

5. A É (A · B)                         2-4 CP

6. ~A Ú (A · B)                                    5 Impl

[5.1.4.] Multiple Uses of CP in a Single Proof.

You can use CP as many times as you need to in a given proof. As your textbook notes, you can use two applications of CP in order to derive a biconditional. For example (this is from exercise 5-1, #14, p.132):

1. D É G                                              p          /\ (D · G) º D

Since the conclusion you need to derive is equivalent to two conditionals:

(D · G) É D     and      D É (D · G)

then you can use CP two times, once to prove each of the conditionals. Then you can easily derive the biconditional you need by applying Equiv to those two conditionals:

1. D É G                                              p          /\ (D · G) º D

2. D · G                                               AP / \ D

3. D                                                     2 Simp

4. (D · G) É D                                     2-3 CP

5. D                                                     AP / \ D · G

6. G                                                     1, 5 MP

7. D · G                                               5, 6 Conj

8. D É (D · G)                                     5-7 CP

9. [(D · G) É D            ] · [D É (D · G)]         4, 8 Conj

10. (D · G) º D                                                9 Equiv

[5.1.5.] Nested Assumptions.

As your textbook explains on pp.127-28, it is possible to have more than one use of CP “active” at the same time. That is, it is possible to assume a premise and then assume a second premise before you have discharged the first assumed premise. In such cases, the second assumption is said to be “nested” inside the first and is therefore called a nested assumption. The following proof, from p.127, illustrates the simultaneous use of different assumptions:

1. C É D                                              p          /\ A É [ B É (C  É D)]

2. A                                                     AP / \ B É (C É D)

3. B                                                      AP / \ C É D

4. C                                                      AP / \ D

5. D                                                     1, 4 MP

6. C É D                                              4-5 CP

7. B É (C É D)                                                3-6 CP

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

Exercise 5-1 (pp.131-132)

·         complete this exercise for next time; as always, check your even-numbered answers, and we’ll cover some of the odds in class.

[5.2.] “Indirect Proofs.”

Review: a contradiction is a sentence that cannot possibly be true. Recall that on your first exam, you were required to use the truth table method to distinguish among sentences that are contradictions, tautologies (which cannot possibly be false) and contingent sentences (which can be either true or false). A sentence is a contradiction when there is no assignment of truth values to its atomic sentences, and therefore no row on the truth table, on which the sentence is true.

Any sentence of the form  p · ~p  is a contradiction. (You can use the truth table method to confirm this).

A valid argument that has all true premises must have a true conclusion. So if you come across a valid argument that has a contradiction as its conclusion, one (or more) of the premises of that argument must be false. In other words, if you can use valid reasoning to move from a set of premises to a contradictory conclusion, then not all of those premises are true.

This suggests another strategy for showing that a given argument is valid:

§  provisionally assume the negation of the claim you are trying to prove, e.g., if you are trying to prove p, then assume ~p;

§  derive a contradiction (any sentence of the form q · ~q) using that negation and the other premises;

§  if you can derive a contradiction, then you will have shown that, given the truth of the original premises of the argument, p must be true; in other words, you will have shown that the argument is valid.

For example:

1.      D Ú E                     p

2.      D É F                     p

3.      ~E                          p          /\ F

In order to prove F, you can assume ~F and then attempt to derive a contradiction:

4. ~F                            AP / \ F

5. ~D                           2, 4 MT

6. E                              1, 5 DS

7. E · ~E                      3, 6 Conj

8. F                              4-7 IP (for “Indirect Proof”)

The general strategy of Indirect Proof (ID): “add the negation of the conclusion of an argument to its set of premises and derive a contradiction.” (133)

NOTE 1: As with CP, you must use arrows and lines to “bracket off” your assumed premise and everything that follows from it from the rest of the argument. (In this online document, I’ve used shading instead, since I cannot easily create the arrows and lines required.)

NOTE 2: “If the conclusion is already a negation, you can just assume the conclusion minus the initial negation sign, instead of assuming a double negation.” (133) This is why there are two versions of this rule [see diagrams on p.133]

Exercise 5-3 (pp.136)

§  complete this exercise for next time; we’ll cover some of the odds in class

Exercise 5-4 (pp.136-37)

§  don’t worry about first doing them without CP or IP, as the instructions indicate-- just do them using IP

§  complete these for next time; we’ll cover some the odds in class

Stopping point for Thursday February 16, For next time:

·         complete exercise 5-1, 5-3, & 5-4

·         go back to chapter 3 to read ch.3:7-9 (pp.74-80).

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