# 7.4: Derivations without Premises

When we discovered the derived weakening rule, we stumbled across the fact that a derivation (or a subderivation) does not have to use all, or even any, of its premises or assumptions. This fact is about to become important in another way. To fix ideas, let me illustrate with the simplest possible example:

0 __| B __ P

1 | __ | A __ A

2 | __ | __A 1, R

3__ | __A⊃A 1-2, ⊃I

The premise, 'B', never got used in this derivation. But then, as I put it before, who ever said that all, or even any, premises **have** to be used?

Once you see this last derivation, the following question might occur to you: If the premise, 'B', never gets used, do we have to have it? Could we just drop it and have a derivation with no premises? Indeed, who ever said that a derivation** has** to have any premises?

A derivation with no premises, which satisfies all the other rules I have given for forming derivations, will count as a perfectly good derivation. Stripped of its unused premise, the last derivation becomes:

1 | __ | A __ A

2 | | A 1, R

3__ | __A⊃A 1-2, ⊃I

(You might now wonder: Can subderivations have no assumptions? We could say yes, except that an assumptionless subderivation would never do any work, for a subderivation helps us only when its assumption gets discharged. So I will insist that a subderivation always have exactly one assumption.)

All right - a derivation may have no premises. But what does a premiseless derivation mean?

Remember that the significance of a derivation with one or more premises lies in this: Any case, that is, any assignment of truth values to sentence letters, which makes all the premises true also makes all of the derivation's conclusions true. How can we construe this idea when there are no premises?

To approach this question, go back to the first derivation in this section, the one beginning with the premise '**B**'. Since the premise never got used, we could cross it out and replace it by any other sentence we wanted. Let us indicate this fact symbolically by writing **X** for the' premise, thereby indicating that we can write in any sentence we want where the '**X**' occurs

0 __| X __ P

1 | __ | A __ A

2 | __ | __A 1, R

3__ | __A⊃A 1-2, ⊃I

For example, for **X** we could put the logical truth, 'AV~A'. Because the result is a correct derivation, any assignment of truth values to sentence letters which makes the premise true must also make all conclusions true. But 'Av~A' is true for all cases. Thus the conclusion, 'A⊃A', must be true in all cases also. That is, 'A⊃A' is a logical truth. I can make the same point another way. I want to convince you that 'A⊃A' is true in all cases. So I'll let you pick any case you want. Given your case, I'll choose a sentence for **X** which is true in that case. Then the above derivation shows 'A⊃A' to be true in that case also.

Now, starting with any derivation with no premises, we can go through the same line of reasoning. Adding an arbitrary, unused premise shows us that such a derivation proves all its conclusions to be logical truths. Since we can always modify a premiseless derivation in this way, a premiseless derivation always proves its conclusions to be logical truths:

A derivation with no premises shows all its conclusion to be logical truth

Armed with this fact, we can now use derivations to demonstrate that a given sentence is a logical truth. For example, here is a derivation which shows 'Av~A' to be a logical truth:

1 | __ | ~(Av~A) __ A

2 | | ~A&~~A 1, DM

3 | | ~A 2, &E

4 | | ~~A 3, &E

5__ | __Av~A 1-4, RD

I devised this derivation by using the reductio strategy. I assumed the negation of what I wanted to prove. I then applied the derived De Morgan and reductio rules. Without these derived rules the derivation would have been a lot of work.

Let's try something a bit more challenging. Let's show that

{[A⊃(B&~C)]&(~BvD)}⊃(A⊃D)

is a logical truth. This is not nearly as bad as it seems if you keep your wits about you and look for the main connective. What is the main connective? The second occurrence of '⊃', just after the ')'. Since we want to derive a conditional with '[A⊃(B&~C)]&(~BvD)' as antecedent and 'A⊃D' as consequence, we want a subderivation with the first sentence as assumption and the second as final conclusion:

1 | __ | [A⊃(B&~C)]&(~BvD) __ A

| |

| | ?

| | ?

| __ | __A⊃D

__ | __{[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) ⊃I

What do we do next? Work in from both ends of the subderivation. The conclusion we want is the conditional with 'A' as antecedent. So probably we will want to start a sub-sub-derivation with 'A' as assumption. At the top, our assumption has an '&' as its main connective. So &E will apply - to give us two simpler conjuncts which we may be able to use. The first of these conjuncts is a conditional with 'A' as antecedent. We are going to be assuming 'A' as a new assumption any way, so most likely we will be able to apply ⊃E. Let's write down what we have so far:

1 | __ | [A⊃(B&~C)]&(~BvD) __ A

2 | | __ | A __ A

3 | | | [A⊃(B&~C)]&(~BvD) 1, R

4 | | | A⊃(B&~C) 3, &E

5 | | | ~BvD 3, &E

6 | | |

| | | ?

| | __ | __D

| __ | __A⊃D ⊃I

__ | __{[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) ⊃I

To complete the derivation, we note that from lines 2 and 4 we can get the conjunction 'B&~C' by ⊃E. We can then extract 'B' from 'B&~C by &E and apply the derived form of vE to 'B' and '~BvD' to get 'D' as we needed:

1 | __ | [A⊃(B&~C)]&(~BvD) __ A

2 | | __ | A __ A

3 | | | [A⊃(B&~C)]&(~BvD) 1, R

4 | | | A⊃(B&~C) 3, &E

5 | | | ~BvD 3, &E

6 | | | B&~C 2, 4, ⊃E

7 | | | B 6, &E

8 | | __ | __D 5, 7, vE

9 | __ | __A⊃D 2-8, ⊃I

10__ | __{[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) 1-9, ⊃I

You might be entertained to know how I dreamed up this horriblelooking example. Note that if, in the last derivation, we eliminated line 10 and the outermost scope line, line 1 would become the premise of a derivation with 'A⊃D' as its final conclusion. In other words, I would have a derivation that in outline looked like this:

__| X __ *P*

| .

| .

| .

__ | __Y final conclusion

But starting with such a derivation I can obviously do the reverse. I get back to the former derivation if I add back the extra outer scope line, call what was the premise the assumption of the subderivation, and add as a last step an application of ⊃I. In outline, I have

| __| X __ A

| | .

| | .

| | .

| | Y

__ | __X⊃Y ⊃I

Looking at the last two schematic diagrams you can see that whenever you have a derivation in the form of one, you can easily rewrite it to make it look like the other. This corresponds to something logicians call the *Deduction Theorem.*

Here is one last application. Recall from chapter 3 that a contradiction is a sentence which is false for every assignment of truth values to sentence letters. We can also use derivations to establish that a sentence is a contradiction. Before reading on, see if you can figure out how to do this.

A sentence is a contradiction if and only if it is false in every case. But a sentence is false in every case if and only if its **negation** is true in every case. So all we have to do is to show the negation of our sentence to be a logical truth:

To demonstrate a sentence,** X**, to be a contradiction, demonstrate its negation, **~X**, to be a logical truth. That is, construct a derivation with no premises, with **~X** as the final conclusion.

Exercise \(\PageIndex{1}\)

7-5. Demonstrate the correctness of the following alternative test for contradictions:

A derivation with a sentence, **X** as its only premise and two sentences, **Y** and **~Y**, as conclusions shows **X** to be a contradiction.

7-6. Provide derivations which establish that the following sentences are logical truths. Use derived as well as primitive rules.

a) (AvB)⊃~(B⊃A)

b) Mv~(M&N)

c) [H⊃(O⊃N)]⊃[(H&O)⊃N]

d) (D⊃B)⊃{D⊃T)⊃[D⊃(B&T)]}

e) (K⊃F)⊃[~F⊃~(K&P)]

f) [(FvG)⊃(P&Q)]⊃(~Q⊃~F)

g) [L⊃(M⊃N)]⊃[(L⊃M)⊃(L⊃N)]

h) [(SvT)⊃F]⊃{(FvG)⊃H]⊃(S⊃H)}

i) (I&~J)v[(J&K)v~(K&I)]

j) {[C&(AvD)]v~(C&F)}v~(A&~G)

7-7. Provide derivations which establish that the following sentences are contradictions:

a) A&~A

b) (Hv~B)&[(~B⊃H)&~H)]

c) [(H&F)⊃C]&~[H⊃(F⊃C)]

d) [~(GvQ)&(K⊃G)]&~(Pv~K)]

e) [K⊃(D⊃P)&[(~KvD)&~(K⊃P)]

f) ~[~(Nv~R)⊃(N≡~R)]

g) (FvG)≡(~F&~G)

h) [~(FvG)v(P&Q)]&~(~Q⊃~F)

i) (A⊃D)&[(A&~B)v(A&~C)]&[(B&~D)v(B&C)]

(Exercise i is unreasonably long unless you use a derived rule for the distributive law. You have really done the work for proving this law in problem 7-ld.

j) (A≡B)=(~A≡B)

7-8. Consider the definition

A set of sentence logic sentences is *Inconsistent *if and only if there is no assignment of truth values to sentence letters which makes all of the sentences in the set true.

a) Explain the connection between inconsistency as just defined and what it is for a sentence to be a contradiction.

b) Devise a way of using derivations to show that a set of sentences is inconsistent.

c) Use your test to establish the inconsistency of the following sets of sentences:

c1) C≡G, G≡~C

c2) FvT, (FvT)⊃(~F&~T)

c3) JvK, ~Jv~K, J≡K

c4) (GvK)⊃A, (AvH)⊃G, G&~A

c5) D=(~P&~M), P≡(I&~F), ~Fv~D, D&J

7-9. Devise a way of using derivations which will apply to two logically equivalent sentences to show that they are logically equivalent. Explain why your method works. Try your method out on some logical equivalences taken from the text and problems of chapter 3. I

chapter summary Exercise

Provide short explanations for each of the following. Check against the text to make sure your explanations are correct, and save your answers for reference and review.

a) Main Connective

b) Primitive Rule

c) Derived Rule

d) Weakening Rule

e) Contraposition Rule

f) De Morgan's Rules

g) Conditional Rules

h) Reductio Ad Absurdurn Rule

i) Derivations without Premises

j) Tests for Logical Truths and Contradictions