6/Formal Proof of Validity

6/FORMAL PROOF OF VALIDITY YolaFile 16.0720 ForClassDiscussionsOnly/Teacher/ArmandLTan/AssociateProfessor/ DepartmentOfPhilosophy/SillimanUniversity

Because the number of truth values multiplies in consonance with the number of distinct variables, it would be difficult to manage arguments involving more than five variables. Logicians, however, developed the method of proof using the rules of inference as proofs for the validity of complex or extended arguments.

Consider the following argument: (Ref. Copi/Cohen, 1990:303-304, No.10 of Ex.IV):

1. If Smith once beat the firemen at billiards, then Smith is not the fireman. (2) Smith once beat the fireman at billiards. (3) If the brakeman is Jones, then Jones is not the fireman. (4) The brakeman is Jones. (5) If Smith is not the fireman and Jones is not the fireman, then Robinson is the fireman. (6) If the brakeman is Jones and Robinson is the fireman, then Smith is the engineer. Thus Smith is the engineer. [O -Smith once beat the fireman at billiards; M -Smith is the fireman; B -The brakeman is Jones; N -Jones is the fireman; F -Robinson is the fire-man; G -Smith is the engineer].

In constructing a proof we follow these simple steps:

1. Write the number before each statement (as we have done above) and symbolize them. 2. Write the last premise and the conclusion on one line. 3. You may draw a horizontal line separating the premises from the proofs.

Symbolically we have

1. O>-M 2. O 3. B)-N 4. B 5. (-M.-N)>F 6. (B.F)>G / G ------

To prove the validity of this argument by matrix test would require us to write 64 rows of TV for each variable, since the argument has six of them. But the method of proof requires no more than two rules of inference to deduce the conclusion of the argument from its premises. By combining 1 and 2, we can infer -M by modus ponens. Again we use MP to deduce -N from the combination of 3 and 4. By Conjunction, we join -M and -N, giving us -M.-N. From -M.-N and the fifth premise, we infer F by modus ponens. Again by Conjunction we join the fourth premise B with F, giving us B.F which we can use with premise 6 so that we can validly -2- infer G, the original conclusion of the argument by modus ponens. This proves the argument valid using MP and CJ as rules of inference to deduce the conclusion of the argument.

The formal proof for this argument is written as

1. O>-M 2. O 3. B>-N 4. B 5. (-M.-N)>F 6. (B.F)>G / G ------7. -M 1,2 MP 8. -N 3,4 MP 9. -M.-N 7,8 CJ 10. F 5,9 MP 11. B.F 4,10 CJ 12. G 6,11 MP

As can be seen here, we write the numbers and the rule of inference used in a single column to the right of each statement. The numbers and the inference rule on the right column serve as explanatory notation on how we got the statement written on the left column. They are the "justifications" for the proof. For instance, on line 7, the numbers 1,2 and MP tell us that we got -M as a valid conclusion from premises 1 and 2 by the rule of modus ponens. In other words, MP allowed us to infer -M from the premises O>-M and O.

This method of proof may be further clarified by providing some partially constructed arguments. For our part, we write the "justification," that is, the numbers and the inference rule that allowed us to infer validly the statement on the left column.

For the following partially constructed proofs, state the numbers and the inference rule that justify the proofs.

1). 1. A>B 2). 1. (P>Q).(R>S) 2. B>C 2. PvR 3. -C 3. (Q>R).(S>T) 4. –A>E /E 4. -R /T ------5. A>C 5. QvS 6. -A 6. RvT 7. E 7. T

-3-

3). 1. A.B 4) 1. A>B 2. B>(E.F) 2. (A.B)>(CvD) 3. -EvC 3. (CvD)>-E 4. C>D / D 4. (A>-E)>F / F ------5. B 5. A>(A.B) 6. E.F 6. A>(CvD) 7. E 7. A>-E 8. C 8. F 9. D s24. The Rule of Replacement. There are arguments whose proofs we cannot construct simply because no inference rules can be used without necessarily replacing the statements with their logically equivalent expressions. Hence, for these arguments, we need the so-called rule of replacement. For this rule, the following tautologies are used.

1. DN (Double Negation) p=(--p) 2. TA (Tautology) p=(pvp)or(p.p) 3. MI (Material Implication) (p>q)=(-pvq) 4. ME (Material Equivalence) (p:q)=[(p>q).(q>p)] =[(p.q)v(-p.-q)] 5. DM (De Morgan Theorem) -(pvq)=(-p.-q) -(p.q)=(-pv-q) 6. TR (Transposition) (p>q)=(-q>-p) 7. EX (Exportation) [(p.q)>r]=[p>(q>r)] 8. CM (Commutation) (p.q)=(q.p) (pvq)=(qvp) 9. AS (Association) [p.(q.r)]=[(p.q).r] [pv(qvr)]=[(pvq)vr] 10. DT (Distribution) [p.(qvr)]=[(p.q)v(p.r)] [pv(q.r)]=[(pvq).(pvr)]

As you can see, these are not arguments (unlike the inference rules) but logically equivalent expressions. Their role in proof construction is to replace (hence replacement) and thereby sim- plify statements so that proofs can be provided using the usual inference rules. The rule of replacement states that "any logical ly equivalent expressions may replace each other whenever they occur”(Copi/Cohen, 1990:304). -4-

1). 1. A>B 2). 1. -(AvB) 2. B>C 2. (-A.-B)>C 3. A>-C /-A 3. -Cv(E:F) /E>F ------4 A>C 1,2 HS 4. -A.-B 1 DM 5. –C>-A 4 TR 5. C 2,4 MP 6. C>-A 5 DN 6. --C 5 DN 7. A>-A 4,6 HS 7. E:F 3,6 DS 8. -Av-A 7 MI 8. (E>F).(F>E) 7 ME 9. -A 8 TA 9. E>F 8 SP

In the first example, we have to replace the third premise A>-C by Transposition (line 5) and by Double Negation (line 6) so that we can proceed with A>C (line 4) and C>-A (line 6)by Hypo thetical Syllogism. These rules help facilitate the construction of proofs.

The second example is more complicated, and as can be observed, no inference rules would allow us to infer validly from the combination of any of the premises. So as the first step, we use the De Morgan Theorem (line 4) without which it would seem impossible to provide proofs for this argument.

Exercises: A. For the following partially constructed arguments, write the number(s) and the rule of inference and replacement.

3). 1. (p.q)>r 4). 1. a>-(bvc) 2. p 2. (-av-b)>c /c 3. (q>r))s ------4. -sv(t>m) /-tvm 3. --(bvc)>-a ------4. bvc)>-a 5. p>(q>r) 5. -(bvc)v-a 6. q>r 6. (-b.-c)v-a 7. s 7. -av(-b.-c) 8. --s 8. (-av-b).(-av-c) 9. t>m 9. -av-b 10. -tvm 10. c

B. Construct a formal proof of validity.

1). 1. a>b 2). 1. a>b 3). 1. av-(bvc) 2. a 2. (-a>-b)>c 2. b /a 3. b>(c.e) 3. -cv(d.e) /d ------4. -cve /e ------

Note: Go to your handout for more Exercises.