[4.] Proofs

 

 

[4.1.] “Argument Forms.” (ch. 4:1)

 

 

argument form: “a group of sentence forms such that all of its substitution instances are arguments” (p.88); for example:

 

p É q

p

\q

 

This is an argument form, since every substitution instance of it is an argument. For example,

 

A É B

A

\ B

 

is a substitution instance of that form, and it is clearly an argument: “A series of sentences, one of which (the conclusion) is claimed to be supported by the others (the premises).” (p.16)

 

The following are also substitution instances of that argument form:

	(E · F) É G E · F \ G

 

(A º B) É ~[B É (C Ú D)] A º B \~[B É (C Ú D)]

           

 

 

 

 

Some argument forms have no invalid substitution instances; i.e., every substitution instance of such a form is a valid argument. Such forms are valid argument forms.

 

An argument form that has even one invalid substitution instance is an invalid argument form. For example,

 

p É q

~p

\~q

 

has as one of its substitution instances

 

A É B

~A

\~B

 

But that argument is invalid; so the form of which it is a substitution instance is invalid.

 

 

 

[4.2.] Review: the Truth Table Method.

 

Much as we can demonstrate that an argument is valid or invalid using the truth table method (as on your first test), we can also use that method to show that a given argument form is valid or invalid.

 

1.       p É q

2.      

p q   p É q p q T T   T T T T T T F   T F F T F F T   F T T F T F F   F F F F F

p

3.      

®

\ q

 

 

 

 

 

There is no substitution instance of this argument that has all true premises and a false conclusion. We can tell this because there is only one interpretation (row 1) on which both premises are true, and on that interpretation the conclusion is true. So there are no interpretations on which all the premises are true and the conclusion false. In other words, the truth of the premises guarantees the truth of the conclusion.

 

--

 

We will use valid argument forms in a new method of demonstrating an argument’s validity. This new method, the proof method, works very much like the method of constructing proofs in geometry. We will demonstrate that a given argument is valid by proving that its conclusion follows from its premises (much as in geometry you would derive a theorem from axioms). 

 

In order to use the proof method, you will need to learn eighteen distinct valid argument forms.  These forms will serve as rules which you will follow in constructing proofs. You must commit all of these forms to memory, just as you did the truth table definitions of the five connectives. We will now begin to introduce those forms/rules.

 

 

[4.3.] Modus Ponens (MP) [Chapter 4:2, p.90]

 

This is the form we examined above:

 

p É q

p          /\q

 

In effect, this says that anytime a conditional is true and its antecedent is true, its consequent must also be true. In other words, if you have a conditional on one line and the antecedent of that conditional on another, then you may write down the consequent of that conditional on a new line.

 

 

Here is how you would employ this rule using the proof method. You’ll be given an argument, as follows:

 

1. B

2. B É ~D                    /\~D

 

You will then construct a proof, showing that the conclusion can be derived from the premises given according to the rules embodied in the eighteen valid argument forms mentioned above:

 

1. B                  p                                  “p” stands for premise

write the abbreviation for the rule, preceded by the numbers of the lines to which you are applying the rule

2. B É ~D        p   /\~D

3. ~D                1, 2 MP                                   

 

 

In this example, we have applied the rule MP (modus ponens) to lines 1 and 2 to derive “~D”, the conclusion of the argument.

 

Notice that it doesn’t matter that in the rule/form, the conditional comes first, whereas in the argument the conditional comes second. As long as you have both the conditional and its antecedent, MP allows you to write down the consequent on a new line.

 

Another proof using MP:

 

1.       A                           p

2.       A É (B v ~C)         p   /\B v ~C

3.       B v ~C                   1, 2 MP

 

Do not confuse MP with the following invalid argument form:

 

p É q

q          /\p

 

This form is called affirming the consequent (notice that in the second line, “q”, the consequent of the opening conditional is being affirmed).

 

 

[4.4.] Modus Tollens [Chapter 4:2, p.92]

 

p É q

~q        /\~p

 

I.e., any time a conditional is true and its consequent is false, its antecedent must also be false.

 

I.e., if you have a conditional on one line and the negation of the consequent of that conditional on another, then you may write down the negation of the antecedent of that conditional on a new line.

 

In a proof (together with MP):

 

1.       D                           p

2.       B É A                    p

3.       D É ~A                  p / \~B

4.       ~A                         1, 3 MP

5.       ~B                          2, 4 MT

 

Do not confuse MT with the following invalid argument form:

 

p É q

~p        /\~q

 

This form is called denying the antecedent (notice that in the second line, “~p”, the antecedent of the opening conditional is being denied).

 

 

[4.5.] Disjunctive Syllogism (DS) [Chapter 4:3 p.93].

 

There are two forms of DS:

 

p Ú q ~p \ q

p Ú q ~q \ p

 

In English, this rule says: if you have a disjunction and the negation of one of its disjuncts, you may write down the other disjunct. (It doesn’t matter which order the disjunction and the negation come in.)

 

Here is an argument, the proof of which uses DS (this one has somewhat complicated premises):

 

1. [~A É (B · C)] Ú (B º ~D)

2. ~(B º ~D)                                         /\~A É (B · C)

 

And here is how to prove that it is valid:

 

1. [~A É (B · C)] Ú (B º ~D)               p

2. ~(B º ~D)                                         p   /\~A É (B · C)

3. ~A É (B · C)                                                1, 2 DS

 

 

[4.6.] Hypothetical Syllogism (HS) [Chapter 4:3 p.93]

 

p É q

q É r    /\p É r

 

This form/rule amounts to the following: if you have two conditionals, and the consequent of one is identical to the antecedent of the other, then you may write down a new conditional with an antecedent identical to that of the first conditional and a consequent identical to that of the second conditional. Here’s HS in a proof:

 

 

 

1.       A É ~B                  p

2.       ~C É A                  p   /\~C É ~B

3.       ~C É ~B                1,2 HS

 

 

Exercise 4-1 (p.94)

§         complete all ten of these at home -- just write down the rule and lines-- no need to write out the whole thing; check your even-numbered answers, and we will review the odds in class next time.

 

Exercise 4-2 (pp.85-86)

§         complete all 14 of these at home; check your even-numbered answers, and we will review the odds in class next time.

 

 

[4.7.] Simplification (Simp). [Chapter 4:4, p.95]

 

Like DS, this rule has two forms:

 

p · q \ p

p · q \ q

 

I.e., if a conjunction is true, then either conjunct is true.

 

I.e., if you have a conjunction, you can “break out” either of its conjuncts and write it on a new line.

 

 

[4.8.] Conjunction (Conj). [Chapter 4:4, pp.95-96]

 

p

q   /\ p · q

 

I.e., given any two sentences, you can join them to make a conjunction.

 

Alternate example of a proof using Simp and Conj (see also Example on p.96):

 

1.       A · B                     p

2.       (A · C) É D           p

3.       C                            p   /\D

4.       A                           1 Simp

5.       A · C                     3, 4 Conj

6.       D                           2, 5 MP

 

 

[4.9.] Addition (Add). [Chapter 4:5, p.96]

 

p   /\ p Ú q

 

I.e., if p is true, then any disjunction in which p is a disjunct is true.

 

I.e., you can take any sentence you already have and use it together with any other sentence (any sentence at all, even if you don’t already have it somewhere else in the proof) to create a disjunction.

 

So from                        A

 

you can derive               A Ú B

 

and                               A Ú ~B

 

and                               A Ú [anything whatsoever]

 

 

Strategy: If the conclusion of your proof contains one or more constants that do not appear in the premises, you can introduce those constants into the proof by using Add. For example:

 

1.       A É (B · C)                        p

2.       D · A                     p   /\C Ú Z

3.       A                           2 Simp

4.       B · C                      1, 3 MP

5.       C                            4 Simp

6.       C Ú Z                     5 Add

 

 

[4.10.] Constructive Dilemma (CD). [Chapter 4:5, p.96-97]

 

p Ú q

p É r

q É s   /\r Ú s

 

I.e., when a disjunction is true, at least one of its disjuncts is true, and if each of those disjuncts materially implies a different claim, then at least one of those different claims is true.

 

I.e., when you have a disjunction and two conditionals, and the antecedents of those conditionals are the disjuncts of that disjunction, you may write down a new disjunction, using the consequents of those conditionals as disjuncts.

 

Note that, unlike with the first seven forms, when you use CD, you must cite three lines (one disjunction and two conditionals).

 

 

1. H É (~G · D)            p

2. H v J                                    p

3. J É (D º C)                          p / \ (D º C) v (~G · D)

4. (D º C) v (~G · D)   1, 2, 3 CD

 

 

[4.11.] Summary of Valid Eight Argument Forms.

 

So far, we’ve seen:

§         Disjunctive Syllogism (DS)

§         Hypothetical Syllogism (HS)

§         Modus Ponens (MP)

§         Modus Tollens (MT)

§         Simplification (Simp)

§         Conjunction (Conj)

§         Addition (Add)

§         Constructive Dilemma (CD)

 

These eight forms are all implicational argument forms. This means two things:

 

1. They are “one-directional.” They embody implications that move only in one direction. For example, using Add, you can reason as follows:

 

A   /\ A Ú B

 

but not as follows:

 

A Ú B   /\ A

 

2. You may apply each of these rules only to entire lines, not to parts of lines.

 

E.g., you cannot apply Add to                 A É B

 

to get                                                    (A Ú Z) É B

 

 

Exercise 4-3:

§         complete all of these at home -- write only what's missing, no need to write entire proof; check your even-numbered answers and we will go through the odds next time.

 

 

Stopping point for Wednesday February 6. For next time, exercises 4-1, 4-2, & 4-3; and read ch.4:6-7 (pp.99-107).