Phil 160: Formal Logic (002)

Notes and examples

Lesson 2

Symbolization is like learning a language. There is a limit to what counts as well formed. We have two kinds of connective - unary (the tilde is the only case) and binary (ampersand, wedge, horseshoe and triple bar.) Whenever we take some sentence and form its negation, we simply add the tilde. No new parentheses are involved. In principle, whenever we take a pair of sentences and form a compound with one of the binary connectives, we put parentheses around the whole.

In the case of a binary sentence that is not being used as a component of something else, we can agree to leave out the parentheses. Otherwise they must be there.

E.g.

~A

never gets parentheses around it even when it becomes part of a larger sentence

(B & ~A)  but not  (B & (~A))

We can drop the outer parentheses like this

B & ~A

unless our sentence becomes a component of a larger sentence, as in

(B & ~A) É ~C
 
 

Examples:

The previous assignment asked you to do 2.1E1 and 5 in lesson 1. These involved relatively straightforward translations. 2.2E1 and 3 ask you to deal with more complex cases. Some advice: the book asks you to "paraphrase". Don't bother unless you find it helpful. Just try to come up with a symbolization. Also, you're not going to be asked to do the reverse - go from SL to English - so you can neglect 2.2E2 and 4.
 
 
 
 

2.2E1

b. At most one of them will win a gold medal.

I.e. at most one of the French, German and Danish teams will win. So we're going to say something involving

F     G     D

'At most one' means 'either one or none' so we need a disjunction that says that one and no more will win or none will win. We'll say that none will win like this

~F & (~G & ~D)

(The positioning of the parentheses was not important.) Now we need to say that one and only one of them will win. We can do this by saying that either the French will win and the others won't, or the Germans will win and the others won't, or the Danes will win and the others won't. Try this

(F & ~(G Ú D)) Ú ((G & ~(F Ú D)) Ú (D & ~(F Ú G)))

Note: '~(G Ú D)' is equivalent to '(~G & ~D)', so either is acceptable. Now we wanted the disjunction between this, which says that one and only one will win and the earlier sentence which says that none will

(~F & (~G & ~D)) Ú ((F & ~(G Ú D)) Ú ((G & ~(F Ú D)) Ú (D & ~(F Ú G))))

Hmm! That's pretty complicated. Here's a slightly simpler version that turns out to be equivalent

(F É ~(G Ú D)) Ú ((G É ~(F Ú D)) Ú (D É ~(F Ú G)))

This says "if the French win, the Germans and Danes won't, and if the Germans win, the French and Danes won't, and so on," but it doesn't say that any of them actually will win.

2.2E3

f. The Germans will win a gold medal only if it doesn't rain during most of the competition and their star runner is not disqualified.

We've got two issues to deal with here. One concerns the main connective - is it 'only if' or 'and'. The other concerns what to do with 'only if'.

First, is the sentence a conditional or a conjunction. I.e., does it tell us two necessary conditions under which the Germans will win - one abut rain and the other about a star runner - or does it tell us one condition - the one about the rain - and then add some extra information? One way to decide is by thinking about which is grammatically or idiomatically better. "The Germans will win only if X and Y" sounds fine, but "the Germans will win only if X. And [by the way] their star runner is not disqualified" is a little awkward.

So let's go with the former. Here are the atomic sentences

G     R     S

We want 'G only if not R and not S'.

G       (~R & ~S)

What about 'only if'? It's the opposite of 'if', so since 'P if Q' goes to '(Q É P)', 'P only if Q' goes to
'(P É Q)'

G É (~R & ~S)
 
 

2.3E1 and 2

 This whole section on truth-functional versus non-truth-functional compounds is extremely thorny. It's more a matter for the philosophy of logic than for logic itself. For the most part, let's leave it well alone. Nevertheless, b and d are useful for illustrating the distinction. I won’t tell you the answers right away, but think about whether the truth value of the compound sentence is determined by the truth values of its components. 2.4E1

d. ‘Copper’ is copper.

False. The first instance, the one in the single quotations refers to the word ‘copper’; the second refers to the substance (or maybe the color.) Thus this sentence says that the word ‘copper’ is itself copper, which is not true.
 

2.4E3

h. (U & C & ~L)

Not an SL sentence. Not too much wrong with it except that each connective must unambiguously bind two other sentences. So if the first ampersand binds 'U' and 'C', then the second must bind '(U & C)' with '~L' or else there will be nothing holding the whole sentence together - no main connective.

2.4E4

d. (K É (~K É K)

The main connective is the first horseshoe. That's what holds everything else together. So the immediate sentential components are

K and (~K É K)

of which 'K' is atomic. The main connective of '(~K É K)' is the horseshoe, so its components are

~K and K

Again, 'K' is atomic.
 

2.4E5

h. ~~(A É B) É (C É D)

In looking for the form ~P É Q, we want a sentence whose main connective is a horseshoe, with the antecedent a negation. In this case, the main connective is a horseshoe, and the antecedent

~~(A É B)

is the negation of

~(A É B)

So this does have the requisite form.

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Last modified 02/10/02
Andrew G. Black
Dept. of Philosophy
UMSL
ablack@umsl.edu