Logical Calculus, Part 2
by Vern Crisler
Copyright 2000; 2005
6. Multitermed
Arguments:
Consider the following argument, one that is typical
of the problems one encounters in the propositional calculus.
1. ~(D v N) É
M
2. (K ·
~A) É C
3. ~(L É
~E) É ~M
4. ~K É
(~D v ~H)
5. ~A É
~(~H · L)
6. ~(H É
M) É B
7. N É
(A v ~B)
8. ~(A É
M) É ~E
Is there an easy way to
solve this argument?
The answer is yes, but before solving it, we need to
remember a fundamental principle of logical calculus: the law of
association. This law tells us that we
can take any combination of alternatives and group them freely. Hence, an expression such as A v B v C can
be combined either as:
(A v B) v C
(C v A) v B
A v (B v C)
And there are more, but the point is obvious: when
dealing with alternatives in any proposition, the variables are free in terms
of their grouping. This goes also for
conjunctions such as A · B · C:
(A · B) · C
(A · C) · B
A · (B · C)
And so on.
Now keeping this principle in mind, we can put the above argument with
eight premises into fractional form, remembering that we can group freely:
1
2 3 4 5 6 7 8
_{}_{}
Because we are able to
group freely, we can go ahead and cross out the polarized terms (e.g., D, ~D, H ~H, etc.). In the case of the third and eighth
premises, the letters M and E are duplicated, but with differing term
operators. The M’s cancel, and the
~E/~E combination reduces to ~E. We end
up after this with three remaining terms: ~L, ~E, and C, as follows:
_{}
Recall that a combination of fractions, with
polarized terms, results in a fraction (or universal conclusion). Converting the fraction to standardized form
and grouping ~L and ~E, we have (~L v ~E) v C, and this is translatable into
~(L · E) v C, and further into (L · E) É C. So “by fell swoop” (as Quine likes to say),
we’ve solved the problem. Interestingly
enough the above eight premises are also translatable into logical algebra, as
follows:
1 

2 

3 

4 

5 

6 

7 

8 
(~d~n)_{a}m 
+ 
(k~a)_{a}c 
+ 
(le)_{e}m 
+ 
(dh)_{a}k 
+ 
(~hl)_{a}a 
+ 
(h~m)_{a}b 
+ 
(~ab)_{e}n 
+ 
(a~m)_{e}e 
And from here, it is
fairly simple to get it back into Lewis Carroll’s original sorites problem,
given in his particular logical notation (or as best as I can render it) on
page 287 of his Symbolic Logic
(Bartley edition):
d´n´_{1}m´_{0
}† ka´_{1}c´_{0 }† le_{1}m_{0 }† dh_{1}k´_{0
}† h´la´_{0 }† hm´_{1}b´_{0 }† a´bn_{0} †
am´_{1}e_{0}
This is what I mean
when I say there is a virtual equivalency between categorical logic and
compound logic at the level of deep structure.
Are we always safe when
duplicated terms are in more than one premiss. Can we just cancel them like we did above with M and E? The answer is no, because in a situation
like this, the middles may not be polarized.
The easiest way to check is to reduce alternating terms to
conjunctions. If the conjunctions
polarize, then the terms can be cancelled as usual. For instance, suppose the following terms were in more than one
premiss:
1^{st}
premiss: ~p v ~q
2^{nd} premiss: pq
It would be possible to
convert the alternating terms of the first premiss to conjunctive form,
resulting in ~(pq). It’s obvious that
the conjunctive forms ~(pq) and pq are polarized, so they can be cancelled as
usual. As long as the situation doesn’t
arise where the same terms in one premiss are positive, and the same terms in
another premiss are negative, then the polarized terms can be cancelled as
usual.
To understand this better, let’s try an
example. Consider the following invalid
argument (from Copi):
1) I É
(J v K)
2) J É
(N É L)
3) ~L / ~I
Here is what it would
look like using fractional notation:
_{}
Or in arithmetical
symbolism:
1. ~I + J +
K
2. ~J + ~K + L
3. ~L
4. no conclusion
In the above argument,
it is important to notice the duplicated terms in premises 1 and 2. We have the same terms in both premises but
with different term operators for each premiss. The terms that are duplicated are J and K. In Copi’s example, the expression J v K is
not polarized with respect to ~J v ~K.
You can see this if we convert the alternations to conjunctions, hence J
v K to ~(~J ·
~K) and ~J v ~K to ~(J · K), i.e.,
_{}
Or,
_{}
While L cancels out, ~I
cannot be derived due to the fact that ~(~J~K) and ~(JK) are not linking
terms. The correct polarization of
~(JK), for instance, is simply JK, not ~(~J~K), and the correct polarization of
~(~J~K) is simply ~J~K.
Again, care needs to be taken that the same terms of
one premiss are not all positive and the same terms of the other conjoined
premiss are not all negative. This
means we can’t cancel them as individual terms, but must test them by conjunctive
form, examining whether the middles are polarized or not. Obviously, the middles are not polarized in
the above argument, and any conclusion derived from it would be invalid. The form j + k conjoined with j + k is (j + k) & (j + k), which is (jj + jk + jk + kk), which can be reduced to
jk + jk. This latter is exclusive disjunction,
meaning “j without k or k without j”—“either the one or the other, but not
both.” (See Boole, Laws of Thought,
for earliest though slightly different example, p. 56.)
The fallacy noted above should not be confused with
the alternation of conjunctions, pq v pq.
The latter is material equivalence, meaning that p=q. Our motto is: alternated conjunctions are
okay but conjoined alternants are not okay¾sometimes. The “sometimes” refers to sameness of signs
with different term operators in conjoined alternants, which should be a
warning bell. Other conjoined
alternants can be okay, however. For
instance, a valid form is p + q conjoined with p + q, that is, (p + q) (q +
p). Notice that the term operators
are different for each premiss, not all positive in one nor all negative in the
other. The latter form can be reduced
to pq v pq, which again is
material equivalence, p=q. Another
example is p + q conjoined with p + q, that is., (p + q) (p + q), which is (pp + pq + pq + qq), which is p + p (q + q) + 0, which is
simplified to p + p (1), or p + p , which is p. Notice that at least one of the premises has a positive and a
negative term operator—so that they are not all positive in one premiss and all
negative in the other premiss. As a
matter of fact, this last example can be found in Carroll’s argument above, in
premises 3 and 8, with terms M and E.
So the bottom line on duplicated terms in more than
one premiss is this. Look out for
sameness of signs and term operators in conjoined alternants. Try to memorize thatconjoined alternants,
conjoined alternants, etc. Close your
eyes and picture in your mind what it looks like. It’s this form: (x + y) & (x + y).
But for me the arithmetical symbolism is the easiest to remember, hence,
1. x + y
2. x + y
The arithmetical forms 1 & 2 present the
illusion that you can just cancel the terms arithmetically, but we have seen
that a simple translation to their conjunctive forms shows that the middles are
not polarized. If the conjoined
alternants have different term operators as noted above, cancel individual
terms in arithmetical fashion. If unsure
about conjoined alternants, try reducing the alternants to conjunctive form and
see if the resulting middles are polarized, then the polarized conjunctions can
cancel one another. Therefore, a “fell
swoop” approach requires care about duplicated terms. For the fellswoopers, just remember that as long as the signs
are not all positive in one alternant and all negative in the other, the
conjunction of these premises will result in a valid conclusion. Did you close your eyes? Can you turn away and picture it in your
mind? If not, better translate
conjoined alternants to their conjunctive forms.
7.
Elementary Valid Arguments Forms
Before moving on to
solving some problems in logical calculus, we should take the time at this
point to state some of the more general laws of logical calculus. These laws can be found in any logic textbook,
but the following discussion symbolizes them using fractional and where
possible arithmetical notation. (Note:
this follows Copi’s presentation.)
1) Modus
Ponens
p É q
p
\q
_{}
1. ~p + q
2. +p
3. 0 +
q
2) Modus
Tollens
p É q
~q
\~p
_{}
1. ~p +
q
2. +~q
3. ~p +
0
3) Hypothetical
Syllogism
p É q
q É r
\p
É r
_{}
1. ~p + q
2. ~q + r
3. ~p + 0 + r
4) Disjunctive
Syllogism
p v q
~p
\q
_{}
1. p + q
2. ~p
3. 0 + q
5) Constructive
Dilemma
(p É q) · (r É s)
p v r
\q
v s
_{}
A constructive dilemma
is translatable to (p v r) É
(q v s), which is then translated into ~(p v r) v (q v s), or (~p · ~r) v (q v s),
which is the same as the above equation.
In arithmetical format, it would be:
1. ~p + q
2. + ~r + s
3. p
+ r
4. 0 +
q + 0 + s
6) Destructive
Dilemma
(p É q) · (r É s)
~q v ~s
\~p
v ~r
_{}
A destructive dilemma is translatable to (~q v ~s) É (~p v ~r), which is then translated into
~(~q v ~s) v (~p v ~r), or (q · s) v (~p v ~r), which is
the same as the above equation. In
arithmetical format:
1. ~p
+ q
2. + ~r + s
3. ~q ~ s
4. ~p + 0 + ~r + 0
7) Simplification
p · q
\p
pq
\p
8) Conjunction
p; q
\p
· q
p; q
pq
9) Addition
p
\p
v q
_{}
8. Logically
equivalent or biunivocal expressions:
1) De
Morgan’s Theorems
a) ~(p ·
q) º (~p v ~q)
_{}
Or, ~(pq) º ~p + ~q.
b) ~(p v q) º
(~p · ~q)
_{}
Or, ~(p + q) º ~p~q
2) Commutation
a) p v q º
q v p
_{}
Or, p + q º q + p
b) (p ·
q) º (q · p)
_{}
3) Association
a) p v (q v r) º
(p v q) v r º
_{}
b) p ·
(q · r) º (p · q) · r
_{}
4) Distribution
a) p ·
(q v r) º
(p · q) v (p · r)
_{}
Or, p(q + r) º pq + pr
b) p v (q ·
r) º (p v q) · (p v r)
_{}
Or, (p + qr) º (p + q)(p + r)
5) Double
negation
_{}
6) Transposition
p É q º ~q É ~p
_{}
Or, (~p + q) º (q + ~p)
7) Definition
of Material Implication
p É q º ~p v q º
_{}
Or, ~p + q.
8) Definition
of Material Equivalence
a) (p º
q) º (p É q) · (q É p)
_{}
(p º q) º
1. ~p
+ q
2. +p +
~q
Notice that the
conjoined alternants have different term operators.
b) (p º
q) º (p · q) v (~p · ~q)
_{}
(p º q) º pq + ~p~q
In this case, we don’t
have conjoined alternants, but rather alternating conjunctions.
9) Exportation
(p · q) É r º p É
(q É r)
_{}
~p + ~q + r
10) Tautology
a) p º
(p v p)
_{}
p º p + p
b) p º
(p · p)
Examples and Solutions to Problems (from Copi, Symbolic Logic, pp. 46ff.):
1)
A É B
~(B v C)
/\
~A
_{}
The conclusion ~A~C is
reduced to ~A by simplification.
1. ~A
+ B
2. ~C & ~B
3. ~A &~C+0
2)
~D v E
~F É ~E
/\
D É
F
_{}
The middle terms are polarized, and at least one of
the “denominators” is positive.
Remember also that ~D v F is easily translated into D É F.
Arithmetically we have:
1. ~D + E
2. ~E + F
3. ~D + 0 + F
3)
(G É H) · (I É H)
J É (G v I)
J
/\
H
_{}
In solving this argument, recall that H v H is a tautology
and is the same as H, so that we could also represent the conclusion as just H.
1. ~G +
H
2. + H + ~I
3. +G
+ I + ~J
4. + J
5. 0
+ H + 0 + 0
4)
(K · L) v (M · N)
K É ~K
/\
M
The following equations
show how a middle or linking term can be “constructed”:
_{}
_{}
_{}
_{}
In solving this
argument, recall that (~K v ~K) º
~K. By means of “addition,” we can construct
a linking term in order to solve the argument, i.e., ~K can add ~L to
give us ~K v ~L, or ~(K ·
L), which is then polarized visàvis KL, and can be cancelled. Also, the conclusion M · N can be
simplified to M.
1. KL + MN
2. ~(KL)
3. 0
+ MN
5.
(O É P) · (Q É R)
(P É S) · (R É T)
S É ~T
O É Q
/\
~O
Finding middle terms in arguments involving
destructive dilemmas takes a bit of “experimentation.” An inspection of the argument shows that we
will have a ~S v ~T in the resulting equation, and in terms of the logic of
destructive dilemmas, this means we’ll also have ~P v ~R, and this will
generate a new destructive dilemma that can be applied to the first premiss in
the above argument. To symbolize it, we
would symbolize the first two premises as destructive dilemmas:
_{}
The combination (PR)
gives us a polarization with ~P v ~R, i.e., ~(PR), and (ST) polarizes
with ~S v ~T, i.e., ~(ST):
_{}
Since a disjunction is freely convertible, we can
translate ~O v Q as Q v ~O, and this will allow us to put the middle terms in
diagonal opposition, making it easier to see how the conclusion is derived:
_{}
The ~O v ~O is a
tautology and is equal to ~O.
1. ~O + P
2. + ~Q + R
3. ~P + S
4. ~R +
T
5. ~S + ~T
6. ~O + Q
7. ~O +
0 +
0 + 0 + 0 + 0
6.
U É (V É W)
(W · X) É Y
~Z É (X · ~Y)
/\
U É (V É Z)
_{}
I tend to avoid a “one fell swoop” approach and I
try to reduce three termed fractions to two termed fractions:
_{}
It is now easy to see which linking terms to
cancel. The conclusion ~(UV) v Z can be
translated into ~U v ~V v Z, and by association, into ~U v (~V v Z), which is
equivalent to U É (V É Z).
1. ~U + ~V +
W
2. ~W + ~(X ~Y)
3. X~Y
+ Z
4. ~U + ~V
+ 0 + 0 + Z
The following are from
Copi’s Introduction to Logic, pp.
338ff, starting with problem 6.
6)
T · (U v V)
T É [(U É (W · X)]
(T · V) É ~(W v X)
/\
W º X
_{}
Or to be more
methodical:
_{}
Remember that by means of association, we can freely
alternate (~W~X) v (WX) to the more regular form of material equivalency, (WX)
v (~W~X).
1. TU
+ TV
2. ~(TU) + WX
3. ~(TV) + ~W~X
4. 0 + 0
+ WºX
Note: The signs WX are positive in one premiss and
negative in the other, but this is only a problem when the variables are
alternants, not conjunctions.
7.
Y É Z
Z É [Y É (R v S)]
R º S
~(R · S)
/\
~Y
_{}
1. ~Y +
Z
2. ~Y + ~Z + ~ (~R~S)
3. ~R~S +
RS
4. + ~(RS)
5. ~Y + 0
+ 0 +
0
8.
A É B
B É C
C É A
A É ~C
/\
~A · ~C
_{}
Solving methodically,
we have
_{}
In the preceding equation, the material equivalence
of A and C allowed us to derive the combination (AC) v (~A~C);
_{}
Combining this with our
remaining fraction, we have:
_{}
Or,
_{}
1. ~A + B
2. ~B + C
3. A + ~C
4. 0
+ 0 + 0
If all the terms cancel, then it’s a sign of
material equivalence between the letters.
In this case, AºC, which is AC + ~A~C. These then are added to the last premiss, as
follows:
5. AC
+ ~A~C
6. ~(AC)
7. 0
+ ~A~C
End of Part 2