Lecture 14: Derived truths + review of TFL

13.2.5 Modus Tollens

Similar considerations as those that lead us to the derived rule of Disjunctive Syllogism

can also be applied to the inference form of Modus Tollens.

Since we can prove (DIY)

(A ! B), ¬B ` ¬A,

we could introduce a derived rule for Modus Tollens:

MT Rule

m A ! B

n ¬B

k ¬A MT m, n

Notice, that with the use of MT, we can also simulate our ! E rule, i. e., Modus Ponens.

13.3 More derived rules

From the example of the derived rules of Disjunctive Syllogism and Modus Tollens,

we see that whenever we have a derivation of the form

P1 , ... , Pn ` Q

we could introduce a derived rule of the following form

Generic derived rule

m 1 P1

...

m n Pn

k Q My New Favorite Rule m 1 , . . . , m n

Of course, it only makes sense to introduce new rules for

sequences of basic inferences that are used frequently in practice.

Otherwise, we’d have to learn and memorize too many rules!

In the textbook, the following derived rules are introduced:

• DS: Disjunctive syllogism

• MT: Modus tollens

• DNE: Double-negation elimination

DNE Rule

m ¬ ¬ A

k A DNE m

• LEM: Excluded middle

LEM Rule

i A

j B

k ¬A

l B

m B LEM i–j, k–l

• DeM: De Morgan Rules

DeM Rules (1)

m ¬(A_ B)

k ¬A^ ¬B DeM m

DeM Rules (2)

m ¬A^ ¬B)

k ¬(A_ B) DeM m

DeM Rules (3)

m ¬(A^ B)

k ¬A_ ¬B DeM m

DeM Rules (4)

m ¬A_ ¬B)

k ¬(A^ B) DeM m

It is a good practice to derive each of these rules!

See Chapter 21 for the rules and also the quick reference (Appendix C) for an overview.

When to use derived rules?

• For learning logic: Use only basic rules! (Like practicing scales on an instrument.)

• For using logic in practice: Use derived rules.

13.4 Brief look back at TFL and a look ahead at FOL

13.4.1 Achievements with TFL

In the TFL that we’ve looked at in the past seven weeks, we have:

• Introduced a (family of) formal languages.

• Clariﬁed the (semantic) notion of validity for these languages, and

made precise and developed an algorithmic test for validity/tautology (truth tables).

• Developed a sophisticated system of (natural) deduction, called ‘Fitch’,

that is based on a few, very straightforward rules of inference.

• We were able to indicate how the deduction system matches validity:

e. g., we saw that it can duplicate DS, Modus Tollens, etc.,

how it can prove ‘ (P ! P) ’, ‘ (P _ ¬P) ’ from no premises, etc.

• Indeed, we can show that for any TFL this system of deduction is sound and complete

with respect to validity/tautology:

– Every sentence that can be proved from no premises is a tautology (Soundness),

– Every tautology can be proved from no premises (Completeness).

All in all, this has been a great success!

13.4.2 Limitations of TFL

However, there are some arguments we use in informal reasoning that cannot be rendered

appropriately in TFL:

1 There are clearly relationships between sentences which TFL is blind to.

Consider, for example, sentences of the kind we began with,

H : ‘Mandy likes hiking’,

S : ‘Mandy likes skiing’,

M : ‘Mandy comes from Montreal’.

Now there is something obviously connecting these sentences,

namely that they are all about Mandy, they all state properties of Mandy.

However, a symbolization of these statements

with the atomic sentence letters ‘ H ’, ‘ S ’, ‘ M ’ clearly doesn’t capture this.

2 There are clearly valid arguments that TFL cannot see as valid.

Consider the argument:

Greta is a hero.

) There is a hero.

This is clearly valid: in any case in which Greta is a hero,

someone is a hero, so there must be a hero.

But its symbolization in TFL is invalid:

) H

13.4.3 Analysis and solution

Analysis of these limitations:

• In TFL we assume that we start from certain sentences (atomic sentences),

and that other sentences of the language are built up from these.

• However, in doing so, we recognize no inner structure of these atoms.

• But sentences like those involving reference to Mandy or Greta,

do have a recognizable inner structure.

Solution:

• To express the inner structure of sentences, we will introduce a family of languages

which are (in various ways) richer (or more complex) than TFL:

they use letters for names and letters for predicates and relations.

These are called ﬁrst-order languages – or, FOL.

After the break!

13.5 More examples of complex proofs [Not presented in

lecture, but read!]

If you have diﬃculties with proofs, here are two detailed walk-throughs of longish proofs.

I would suggest to ﬁrst try to do these proofs by yourself,

reﬂecting on your strategy and reasoning processes!

These proofs are long, so take your time.

1. ` (A ! (B ! C )) $ ((A ^ B) ! C )

2. ` ¬(A $ B) $ (A $ ¬B)

After that, study the rest of this section carefully at your own pace!

13.5.1 Show: ` (A ! (B ! C )) $ ((A ^ B) ! C )

This statement contains quite a few connectives, which makes it not so easy to prove.

Let’s see whether we can apply the general strategy introduced in the previous lecture

to make some progress.

1. The main connective in the formula we want to prove is a biconditional.

There are no premises and there are no disjunctions whatsoever in our goal:

this more or less rules out _ I, so our last step will be either $ I or IP.

2. If we pick $ I, we will need two subproofs,

each of which introduce a conditional (one for each direction).

3. In the ﬁrst subproof, we need to assume (A ! (B ! C )) and derive ((A ^ B) ! C ) .

The goal is a conditional:

so we’ll assume the antecedent (A ^ B) in a new subproof and

easily show the consequent C !

4. In the second main subproof for $ I, we assume the right hand side and

derive the one on the left:

We assume ((A ^ B) ! C ) and derive (A ! (B ! C )) .

Since the goal here is again a conditional, we assume the antecedent A

in a new subproof to derive the consequent (B ! C ) .

This is again a conditional, so we’ll need a third subproof to derive it!

5. That’s the plan. Now we put it into our proof system.

It is important not to get lost in the details.

It can be useful to draw up a sketch of the plan,

or try out a few moves with pen and paper before going to carnap.

From left to right of the biconditional:

1 A ! (B ! C )

2 A ^ B

3 A ^E 2

4 B ! C !E 1, 3

5 B ^E 2

6 C !E 4, 5

7 (A ^ B) ! C !I 2–6

From right to left of the biconditional:

8 (A ^ B) ! C

9 A

10 B

11 A ^ B ^I 9, 10

12 C !E 8, 11

13 B ! C !I 10–12

14 A ! (B ! C ) !I 9–13

Combining the two subproofs:

15 (A ! (B ! C )) $ ((A ^ B) ! C ) $I 1–7, 8–14

13.5.2 Show: ` ¬(A $ B) $ (A $ ¬B)

1. Here, again, the main connective of our goal is a biconditional:

So our options are $ I, _ E, and IP.

2. There are no disjunctions in this sentence, so probably not _ E.

It could be IP, but there doesn’t seem to be an immediate path to derive a contradiction.

3. Thus, our ﬁrst bet should be to use $ I as our last rule.

This requires two subproofs:

(a) one from the left side of the biconditional to the right side, and

(b) one from the right to the left.

4. To derive the left side seems fairly straightforward, using ¬ I:

assume (A $ B) and derive a contradiction.

5. The other direction looks more diﬃcult.

The ‘antecedent’ is a negated biconditional, and the ‘consequent’ is a biconditional.

So we need to introduce a biconditional (which means again two subproofs),

but we also must be able to derive a contradiction. . .

Note: This is one of those cases where an intuitive strategy (say, showing some equiv-

alent version via MT and DeM) will be technically feasible but extremely long. The

shortest way seems to be to build up a ‘dummy’ biconditional that contradicts the

‘antecedent’. It’s long, but doable!

Here’s the ﬁnal proof.

First main subproof, from left to right of the biconditional,

itself using two subproofs in lines 2–10 and 11–21:

1 ¬(A $ B)

2 A

3 B

4 A

5 B R 3

6 B

7 A R 2

8 A $ B $I 4–5, 6–7

9 ? ¬E 1, 8

10 ¬B ¬I 3–9

Second subproof within ﬁrst main subproof:

11 ¬B

12 ¬A

13 A

14 ? ¬E 12, 13

15 B X 14

16 B

17 ? ¬E 11, 16

18 A X 17

19 A $ B $I 13–15, 16–18

20 ? ¬E 1, 19

21 A IP 12–20

22 A $ ¬B $I 2–10, 11–21

Second main subproof, from right to left of the biconditional:

23 A $ ¬B

24 A $ B

25 B

26 A $E 24, 25

27 ¬B $E 23, 26

28 ? ¬E 25, 27

29 ¬B ¬I 25–28

30 A $E 23, 29

31 B $E 24, 30

32 ? ¬E 29, 31

33 ¬(A $ B) ¬I 24–32

Putting together the two main subproofs: