4/ARGUMENT FORMS AND TEST OF VALIDITY 16.0720 ForClassDiscussionsOnly/Teacher/ArmandLTan/AssociateProfessor/
PhilosophyDepartment/SillimanUniversity
s18. Basic Syllogism= an argument drawn fromtwo premises by deductive inference. 1. Disjunctive syllogism (DS).
(1) Either Venn loves April or Venn loves May.
Venn does not love April. Hence, Venn loves May.
(2) Either Venn loves April or Venn loves May.
Venn loves April. Hence, Venn does not love May.
Of these two, the 1STargument, pvq and –p /q, is a valid form while the 2nd, pvq and p /–q, is invalid.
If an argument is valid, then any argument having that same form is also valid. Hence, the arguments of the form
(3) (A.B) v C and (4) C v (A.B)
-(A.B) -C
/ C / (A.B)
arevalid since they followthe same pattern of reasoning or being the substitution instances of thedisjunctive syllogism (1).Since the 2ndargument form is invalid, any argument of this form is also invalid. Thus the disjunctive arguments of the form
(5) (A:B) v C and (6) C v (A>B)
(A:B) C
/ -C / -(A>B)
are invalidbeingthe substitution instances(2). However, if the disjunctive premise is understood to be a strict disjunction, then these are valid forms.
Pure hypothetical arguments.
2. Hypothetical Syllogism (HS): p>q and q>r /p>r
Example: If Christ is God, then He is the only way.
If He is the only way, then believers are saved.
So if Christ is God, then believers are saved.
Mixed hypothetical arguments.
3. Modus Ponens (MP): p>q and p /q
Example:If Salonga is president, then "the weak shall be
strong". Salonga is president.
Hence, the weak shall be strong.
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4. Modus Tollens (MT): p>q and–q / -p
Example: If Salonga is president, then the "strong shall
be just." The strong are not just.
Hence, Salonga is not the president.
Invalid forms of mixed hypothetical argument:
a. fallacy of affirming the consequent-> p>q and q /p
b. fallacy of denying the antecedent -> p>q and –p /-q
5. Conjunction [CJ] 6. Simplification [SP]
p p.q
q /p.q /p
7. Addition [AD] 8. Absorption [AB]
p p>q
/pvq /p>(p.q)
9. Constructive Dilemma [CD] 10. Destructive Dilemma[DD]
(p>q).(r>s) (p>q).(r>s)
pvr -qv-s
/qvs /-pv-r
s19. Matrix Test of Validity: The TT that shows True-True-False or 110 means that the argument tested is invalid.
Example: p q | pvq p -q pvq -p q p>q -q -p
------
r1 1 1 | 1 1 0 1 0 1 1 0 0
1 0 | 1 1 1 1 0 0 0 1 0
0 1 │ 1 0 0 1 1 1 1 0 1
0 0 │ 0 0 1 0 1 0 1 1 1
------
Invalid valid valid
1. HS 2. invalid
p q r│ p>q q>r p>r p q r│ p>q q>r r>p
------
1 1 1│ 1 1 1 1 1 1│ 1 1 1
1 1 0│ 1 0 0 1 1 0│ 1 0 1
1 0 1│ 0 1 0 1 0 1│ 0 1 1
1 0 0| 0 1 0 1 0 0| 0 1 1
0 1 1│ 1 1 1 / 0 1 1│ 1 1 0
0 1 0│ 1 0 1 0 1 0│ 1 0 1
0 0 1│ 1 1 1 / 0 0 1│ 1 1 0
0 0 0│ 1 1 1 0 0 0│ 1 1 1
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s19. Short-Cut: method of assigning TV to make premises true and conclusion false. If successful, then argument is invalid.
Example:
1.p>q -p -q
0>1 1 0
/1
2.(p>q).(r>s) pvr / qvs
In order to falsify the conclusion, we assign 0 to both q and s. Assigning 0 to both q and s necessitates the assignment of 0 also to p and r making the first premise true. But by signing 0 to both p and r makes the second premise false. And if we assign 1 to either p or r making the second premise true, this inturn will make the first premise false. In other words, no matter whatthe TV we assign to both p and r, we can never get 110, i.e. there is no way we can falsify the conclusion of a valid argument.
s20. Tautology and Validity. We have shownthatavalid argument provides no instances of true premises and a false conclusion regardlessofthe truth conditions ofits componentvariables.This observation points to an important relationshipbetweentautologiesthe validity of any given argument.
Any given tf-argument can be translated intoan implica-
tion.
Example: pvq and –p, /q Translation: [(pvq).-p]>q
Matrix: pq [(pvq). -p]>q
---|------
1 1│ 1 0 0 1 1
1 0│ 1 0 0 1 0
0 1│ 1 0 1 1 1
0 0│ 0 0 1 1 0
= tautology /argument is valid.
For invalid arguments the matrix result will either be contingentor self-contradictory.
Example: p>q and q>r /r>p
Translation: [(p>q).(q>r)]>(r>p)
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Matrix: p q r [(p>q).(q>r)] > (r>p)
------|------
1 1 1 | 1 1 1
1 1 0 │ 0 1 1
1 0 1 │ 0 1 1
1 0 0 │ 0 1 1
0 1 1 │ 1 0 0
0 1 0 │ 0 1 1
0 0 1 │ 1 0 0
0 0 0 │ 1 1 1
=contingent /invalid
s18/20 Exercises
I.Determine the invalidity of the following arguments by short-cut method.
1. a=(b.c) 2. p>(q.r) 3. k.h
b=(evf) (q.r)>k fv(k>t)
-c /ave pvr /q>k -f /h
a b c e f k p q r f h k t
------
? ? ?
4. av(b>c) 5. f>(g.h) 6. (l.m)v(o=p)
b>c (g.h)>i (-mvx)>y
-(bvd) jvk y=o
e=b /c -f /-iv-k -(l.m) /pvy
a b c d e f g h i j k l m o p x y
------
? ? ?
II. Translate the following arguments into conditional statements and determine their validity by matrix.
1. p>(q.r) 2.(p.q)v(r.s) 3. pv(q>r)
p /q.r -pv-s /-qv-r -r /-pv-q
4. pv(q.r) 5. (p.q)v(r.s) 6. p>(-pvr)
pvr -q>-s -(-pvr)
/pvq /-p>-r /-p