Logical Calculus, Part 3
by Vern Crisler
Copyright 2000; 2005
9. Simplification
If we want to simplify
the following, how would we do it?
Besides the above rules, there are
rules of logical calculus that are helpful in simplifying complex expressions. We will give them in both linear and
fractional form:
1. (a + b) (a + b)
= a + (bb)
= a
1*
_{}
2. (ab) + (ab)
= a(b + b) = a(1) = a
2*
_{}
3. a + ab = a
3*
_{}
4. a(a + b) = aa
+ ab = a + ab = a
4*
_{}
5. a + ab
= a + b
5*
_{}
6. a(a
+ b) = aa + ab = 0 + ab = ab
6*
_{}
Fractional representation
of premises helps us understand the process of simplification a bit more than
the usual linear representations of simplifications. Here is an example from Keynes, Formal Logic, p. 533:
abc 
abc 
abc 
abc 
Simply as follows:
1 2 3 4
5 6
_{}
The fourth
simplification could be solved in a different way if the “c” is retained and
the “c” is removed,
as follows:
3 4*
5* 6* 7 8
9 10 11
_{}
Keynes’ Formal Logic,
p. 477, asks us to show the equivalency between two expressions:
bc or bd or cd is equal to bc or bd
This is a
simplification problem, and we can solve it fractionally as follows:
_{}
Thus, our solution _{} is equivalent to Keynes’ solution “bc or bd.”
10.
Invalidity
We will repeat the
rules for Logical Calculus, and then give examples of arguments that are
invalid due to a violation of these rules:
1) A fractional expression combined with
a fractional expression leads to a fractional conclusion. (The middle terms must be polarized; cannot
have two negative denominators.)
2) A fractional expression combined with
a conjunctive expression leads to a conjunctive conclusion. (The middle terms must be polarized.)
3) A conjunction combined with a conjunction
results in an invalid conclusion, since two particular statements fail to
distribute the middle terms.
4) A fractional expression combined with
a fractional expression, with like middle terms, may result in a
valid conclusion but only if conversion per accidens is able to polarize
the middles; otherwise, the argument is invalid.
AAO1
M É P
S É M
\S
· ~P
_{}
As you can see, an argument form involving AAO in
the first figure is invalid because it violates rule 1: A fraction combined with a fraction, with
polarized terms, results in a universal (fractional) conclusion, not a
conjunctive conclusion. The conclusion
should have been ~S + P, or S É P.
AIA1
M É P
S · M
\S
É P
_{}
In this case, the conclusion should have been a
conjunction of S and P, as per Rule 2.
Note that the middle terms are polarized. In arithmetical form, we have:
1. ~M + P
2. M & S
3. 0+ S & P
AOA1
M É P
S · ~M
\S
É P
_{}
In this case, the conclusion is neither S É P nor S · P. It cannot be a universal conclusion due to
the fact that it would violate rule 2; but secondly, it cannot be a conjunctive
conclusion due to the fact that the middle terms are unpolarized. In a fraction combined with a conjunction,
the middle terms must be polarized or no conclusion can be validly
derived. (Conversion per limitation
does not work here since it can only be applied to an argument with two
universal premises, not just one.)
1. ~M + P
2. ~M & S
3. xxxxxx
EEO1
M É ~P
S É ~M
\S
· ~P
_{}
Here we have two universal premises but conversion
by limitation cannot be used because both premises have negative
denominators. In logical algebra it
would be: MeP+SeM=? which immediately shows its invalidity (unpolarized
middles).
1. ~M + ~P
2. ~M + ~S
3. xxxxxxxxxxxx
AAA2
P É M
S É M
\S
É P
_{}
The middle terms are
unpolarized.
1. ~P + M
2. M + ~S
3. xxxxxxxxx
11. Logical Calculus & Logical Algebra:
I had originally proposed what I called the Dangling
Operator Rule as a means for testing validity.
I now regard this rule as unnecessary.
I formulated this rule because it seemed that some of the equations I
tested could not pass the rules developed in my essay “Logical Algebra.” Here is an example:
OAI2
P · ~M
S É M
\S
· P
_{}
The two conclusions are
different. If we translate this into
logical algebra, here is what we have:
PoM + SaM = ?
As you can see, neither the first or second premises
distribute the middle term, and if I change PoM to its first figure form (PoM;
Pi~M; ~MiP; ~Mo~P), we would have unpolarized middle terms:
~Mo~P + Se~M = ?
Here is another
example:
AEO1
M É P
S É ~M
\S
· ~P
_{}
MaP + SeM = ?
As you can see, this argument would be invalid in
the first figure since the middle terms cannot be polarized.
After thinking about this
problem for a while, it dawned on me that the equations in logical calculus are
different from the equations in logical algebra in an important respect,
namely, that in logical calculus, the equations are indifferent
with respect to normal form and transposed form. Thus, while some of these equations may be invalid if interpreted
as normal form equations in logical algebra, they could very well be valid if
interpreted as transposed forms in logical algebra. For instance, in the OAI2 form above, the equation:
PoM + SaM = ?
was considered invalid in terms of its normal form,
for we could not achieve a polarization of the middle terms when we tried to
reduce it to the first figure in normal form.
But let us consider what happens when we interpret the equation in terms
of transposed form:
PoM + ~MeS = PoS
Here we have the M’s adjacent to the plus sign, and
read from left to right to derive the conclusion. The conclusion PoS is obvertible to Pi~S, and convertible to
~SiP, which is the ~S & P of the original equation solved in logical
calculus. Thus, in logical calculus the
argument is valid, and this is because logical calculus is indifferent with
respect to normal or transposed form.
(Of course, the conclusion of OAI2 is still invalid, since its conclusion
S & P is not equivalent to the correctly derived ~S & P.)
So also with the second example, AEO1. We were not able to derive a valid
conclusion from the equation MaP + SeM, and this was again due to the fact that
the middle terms could not be polarized.
Again, let us consider what happens if we interpret the equation in
terms of transposed form. The resulting
equation would be:
Po~M + MeS = PoS
Here, the middle terms
are polarized, and the PoS conclusion is obvertible to Pi~S and convertible to
~SiP, which is the ~S & P of the original equation in logical
calculus. (Nevertheless, AEO1 would
still be invalid since the O conclusion S & P
is not ~S & P.)
In sum then, there is
no need to add any new rule of invalidity to the ones listed above. If a fractional and a conjunctive expression
are combined, and the middle terms are polarized, the argument is valid in
logical calculus. If logical algebra is
used to test these fractional/conjunctive equations with polarized middles, the
reader should remember that while a logical calculus equation might be invalid
in logical algebra’s normal form, it could very well be valid in logical
algebra’s transposed form, for the equations of logical calculus
represent either form, or that is to say, are indifferent with respect to
normal or transposed form.
12. Solving the Sorites in Logical Calculus
In our second essay,
“The Sorites,” we discussed the sorites from the stand point of logical
algebra. In this section, we will
discuss how to solve soriteses by means of logical calculus. It is actually a fairly simple procedure,
perhaps even simpler than what we did under logical algebra.
Before providing an
example, let us adopt the following convention for the use of the & sign of
conjunction or particularity. The
statement “Some S is P” is symbolized as S & P, but could just as easily be
symbolized by S& &P, that is,
“Some S is some P.” If we want to place
the premiss (S & P) in a logic grid, we can combine the “&” with either
letter, or with both of them at the same time.
Hence, S& can be in one square of the logic grid, and &P could
be in another square. The reason is so
that we don’t lose track of particularity, especially if one of the particular
terms is cancelled. A particular premiss
means that our conclusion must be particular.
If one is inclined to forget where particular premises are, it might be
a good idea to write a “&” next to the row of the logic grid where the
particular was found. In any case, the
“&” is treated just like a positive value for cancellation purposes. If “&P” is conjoined with “P” it will be cancelled just
as though it were a P standing by itself conjoined with a P.
Another convention we would like to adopt for logic grids
is that the vertical lines (going up and down) represent alternation, while the
horizontal lines (going across) represent conjunction. Here is an example (from Sommers) that
illustrates the above conventions:
1. If some
clergymen are priests, and
2. All
priests are bachelors, and
3. No
bachelors are married, then
\
Some clergymen are not married.
This can be represented arithmetically as
1. C & P
2. – P
+ B
3. –B + – M
\ C & 0 + 0 + – M
Conclusion C & –M, “Some clergymen are not
married.” The following is a logic grid
showing the cancellation procedure for logical calculus.

B 
C 
M 
P 
1 

C & 

& P 
2 
B 


P 
3 
B 

M 

\ 
0 
C & 
M 
0 
The numbers refer to the
premises, and the top row of letters refers to the column where the variables
will be placed. This is for convenience
when doing large logic grids. As can be
seen, placing a “&” next to both variables is a convenient way to keep
track of the logical connective for particulars. As we said in our essay on “The Sorites,” care needs to be taken
that in conjoining a particular premiss with another premiss in the sorites, a
distribution of the middles must take place.
If there is no distribution, the sorites will be invalid.
It should be noted that
placing the “&” in the grid only has reference to logical connectors of the
premiss itself, not to the logical connecters inside a term within the
premiss. For instance, the first premiss
C & P has a logical connective expressing conjunction or
particularity. This particularity holds
between the terms of the premiss and is expressed by the “&” coming
between the terms. However, another
premiss might contain the conjunction sign “&” inside of the terms rather then
between them. Here is an example:
“All –B–C is DE or D–F or –H–K.”
I will repeat the sentence using parentheses to mark
off the logical connectives inside the terms, and bolditalic type to mark off
the logical connectives of the premiss itself:
“All (–B & –C) is
(D & E) or (D & –F) or (–H & –K).”
Here the logical connectives are “All is,” along
with the two “or’s.”
As we said, to determine whether to place a “&”
in a logic grid, it is only the logical connectives of the premises that need
to be taken down to the conclusion, not the connectives making up terms within
each premiss. This will require a
different sort of grid. In the case
below, the grid boxes are not representing alternation, but just recording the
existence of the variables in question.
It’s only the last row (the conclusion) that will represent
alternation. This example is taken from
John Keynes, Formal Logic, p. 524:

Original premises 
Logical calculus 
Grid equations 
1. 
All –B–C is DE or D–F or –H–K 
–(–B–C)
+ DE + D–F +–H–K 
+B
+ C + DE +D–F + –H–K 
2. 
All
C is –AB or DEFG or BFH 
–(C)
+ –AB + DEFG + BFH 
–C
+ –AB + DEFG + BFH 
3. 
All
B–C–D is –EL or –H–K 
–(B–C–D)
+ –EL + –H–K 
–B+C+D
+ –EL + –H–K 
4. 
All
A–C–F is –D 
–(A–C–F)
+ –D 
–A+C+F
+ –D 
5. 
All
–K is BC or C–D or C–F or H 
–(–K)
+ BC + C–D + C–F + H 
K
+ BC + C–D + C–F + H 
6. 
All
ABCDEFG is H or K 
–(ABCDEFG)
+ H + K 
–A+
–B + –C + –D + –E +–F + –G + H + K 
7. 
All
DEFGH is B 
–(DEFGH)
+ B 
–D
+ –E+ –F + –G + –H + B 
8. 
All
AB–L is –F or –H 
–(AB–L)
+–F + –H 
–A
+ –B + L + –F + –H 
9. 
All
ADFK–L is H 
–(ADFK–L)
+ H 
–A
+ –D + –F + –KL + H 
10. 
All
ADEFH is B or C or G or L 
–(ADEFH)
+ B + C + G + L 
–A
+ –D + –E + –F + –H + B + C + G + L 
As can be seen, all of
Keynes’ premises are universal in form, and so the conclusion will retain its
universality. The terms within the
premises such as (DE)
or (–H–K) are particular in form, but as we said they do not
govern the quantity of the conclusion.
Nevertheless, since we still have conjunctive terms such as DE or BC or
DEFG, etc., we cannot use the grid to represent logical connectives. The slots are just placeholders for the
individual letters. If we started
placing conjunctions in the letter slots, the grid would become large and would
serve little purpose. Thus, in a grid
of this sort, we have to forgo the idea that the vertical lines represent
disjunction. They do not only in this
case, i.e., only when the terms of a premiss contain more than one
variable. Here is the logic grid for
the above equations:

A 
B 
C 
D 
E 
F 
G 
H 
K 
L 
1 

B 
C 
D 
E 
F 

H 
K 

2 
A 
B 
C 
D 
E 
F 
G 
H 


3 

B 
C 
D 
E 


H 
K 
L 
4 
A 

C 
D 

F 




5 

B 
C 
D 

F 

H 
K 

6 
A 
B 
C 
D 
E 
F 
G 
H 
K 

7 

B 

D 
E 
F 
G 
H 


8 
A 
B 



F 

H 

L 
9 
A 


D 

F 

H 
K 
L 
10 
A 
B 
C 
D 
E 
F 
G 
H 

L 

A 
0 
0 
0 
0 
0 
0 
0 
0 
L 
To derive the conclusion
on the last row: Check all the premises
to see if A
and L occur together as part of a conjunction.
If not, then they can be treated as disjunctive terms, i.e., –A + L,
which is “If A then L,” or Keynes’ “All A is L.”
Here is a problem from Carroll
(Symbolic Logic, p. 134.):
1. No a are –b
(that is, All a are b)
2. All b are c
3. All c are d
4. No –e are –a (that is, All e are a)
5. All h are –e
In logical algebra we
would symbolize the first premiss as Ae~B, or AaB, and so on for the remaining
premises. In logical calculus, these
are symbolized as follows:
1. A É
B
2. B É
C
3. C É
D
4. –E É
A
5. H É
E
By arithmetical format,
it is easy to derive the conclusion:
1. –A + B
2. –B + C
3. – C + D
4. +A + E
5. –E + –H
6. 0 + 0
+ 0 + D + 0 + –H
The conclusion is D +
–H, which becomes –D É
–H, which can be resolved into H É
D, which is Carroll’s conclusion “All h are d.”
Carroll provides a
longer sorites (Ibid., page 136):
1. All k are l
2. No d are –h
3. All a are –e
4. All b are e
5. No –k are h
6. No –b are l
7. All –d are c
Translating this into
the symbolism of logical calculus, we have:
1. k É
l
2. d É
h
3. a É
–c
4. b É
e
5. –k É
–h
6. –b É
–l
7. –d É
c
Instead of using an
arithmetical format to solve this sorites, we shall use another version of
a logic grid, this time without the
convenient letters across the top. This
grid is small enough so that their absence is not too painful:
1 
–k 
l 






2. 


–d 
h 




3. 




–a 
–c 


4. 






–b 
e 
5. 
k 


–h 




6. 

–l 




b 

7. 


d 


c 


\ 
0 
0 
0 
0 
–a 
0 
0 
e 
Since we only have
single terms, we can treat the above grid as representing logical connectives
throughout. This means that the disjunctive
conclusion of –a + e follows because neither occurs in a conjunction in the
premises. The conclusion can also be
rendered as a É
e, or as Carroll says, “All a are e.”
In the second essay on logical algebra, we presented a sorites given by Carroll (Ibid., p. 305) and showed how it could be solved using logical algebra. In this case, we want to solve the same sorites below by using the symbols and operations of logical calculus along with a logic grid. Carroll’s sorites was as follows:
1 
2 
3 
4 
5 
6 
Cl_{1}E´_{0}
† 
Av´_{1}D_{0}
† 
k_{1}m´_{0}
† 
lC´_{1}(b´n´)´_{0}
† 
dsb_{1}t´_{0}
† 
tD_{1}w´_{0}
† 






7 
8 
9 
10 
11 
12 
dr´a´_{1}A´_{0}
† 
vw_{1}B_{0}
† 
em´_{1}(r´b´)´_{0}
† 
Ha_{1}c´_{0}
† 
dtmav´_{0} † 
dst_{1}A´_{0}
† 






13 
14 
15 
16 
17 
18 
Dn´r´b´_{1}z_{0}
† 
cE´z_{0} † 
bs´_{1}l´e´_{0}
† 
atE_{1}v´_{0}
† 
rDh´_{1}e´_{0}
† 
mt´_{1}D_{0}
† 






19 
20 
21 
22 
23 
24 
Anl´_{1}c´_{0
} † 
rdk´_{1}h_{0}
† 
ztB´_{1}d_{0}
† 
nl´_{1}H´_{0}
† 
Et´_{1}z_{0}
† 
dzrA´_{1}a´_{0} 
The following table shows
the steps I’ve used in getting from Carroll’s symbolism to logical calculus,
and then from there to arithmetical form.
It is not necessary to go through all these steps. It’s possible to go from Carroll’s symbolism
straight to the grid, but solving a sorites too quickly invites
carelessness. My motto—just to state
the obvious¾is
that it is better to be thorough and right than it is to be quick and wrong:

Hypothetical 
Disjunctive 
NeoBoolean 
1. 
[C l] É E 
[C
l] + E 
C
+ l + E 
2. 
[A v] É –D 
[A
v] + –D 
A
+ v + D 
3. 
k É m 
k
+ m 
k
+ m 
4. 
[l C] É –[b n] 

[l C] + –[b n] 
l
+ C + –b + n 
5. 
[d s b] É t 

[d s b] + t 
d
+ s + b + t 
6. 
[t D] É w 

[t D] + w 
t
+ D + w 
7. 
[d r a] É A 

[d r a] + A 
d
+ r + a + A 
8. 
[v w] É B 

[v w] + B 
v
+ w + B 
9. 
[e m] É [r b] 

[e m] + [r b] 
e
+ m + r
+ b 
10. 
[H a] É c 

[H a] + c 
H
+ a + c 
11. 
[d t m a] É v 

[d t m a] + v 
d
+ t + m + a + v 
12. 
[d s t] É A 

[d s t] + A 
d
+ s + t + A 
13. 
[D n r b] É z 

[D n r b] + z 
D
+ n + r + b + z 
14. 
[c E] É z 

[c E] + z 
c
+ E + z 
15. 
[b s] É [l e] 

[b s] + [l e] 
b
+ s + l + e 
16. 
[a t E] É v 

[a t E] + v 
a
+ t + E + v 
17. 
[r D h] É e 

[r D h]
+ e 
r
+ D + h + e 
18. 
[m t] É D 

[m t] + D 
m
+ t + D 
19. 
[A n l] É c 

[A n l]
+ c 
A
+ n + l + c 
20. 
[r d k] É h 

[r d k]
+ h 
r
+ d + k + h 
21. 
[z t B] É d 

[z t B]
+ d 
z
+ t + B + d 
22. 
[n l] É H 

[n l] + H 
n
+ l + H 
23. 
[E t] É z 

[E t] + z 
E
+ t + z 
24. 
[d z r A] É a 

[d z r A]
+ a 
d
+ z + r +A + a 
If we transfer the last
column to a grid, we can solve the sorites by a simple process of
elimination. The numbers 1 through 24
are, of course, the number of the premises.
The letters going across the top, as we said, are for convenience. They help guide the placement of the letters
in the correct row of the premises. The
main problem to watch out for in constructing such a grid is that it is easy to
become careless, and to leave out one or more of the letters. The best way to overcome this problem is
that once the grid is finished, look at the last row containing the conclusion,
and review each remaining variable against the original premises to make sure
all of the polarizations have been taken into account. Also, examine the letters of the conclusion
to determine whether any occur in a conjunction in the premises. Since there were no conjunctions in the
NeoBoolean column, the conclusion will be in disjunctive form.

A 
B 
C 
D 
E 
H 
a 
b 
c 
d 
e 
h 
k 
l 
m 
n 
r 
s 
t 
v 
w 
z 
1 


C 

E 








l 








2 
A 


D 















v 


3 












k 

m 







4 


C 




b 





l 

n 






5 







b 

d 







s 
t 



6 


















t 

w 

7 
A 





a 


d 






r 





8 

B 

















v 
w 

9 










e 



m 

r 





10 





H 


c 













11 






a 


d 




m 



t 
v 


12 
A 








d 







s 
t 



13 



D 



b 







n 
r 




z 
14 




E 



c 












z 
15 







b 


e 


l 



s 




16 




E 

a 











t 
v 


17 



D 





d 
e 
h 




r 





18 



D 





d 




m 



t 



19 
A 







c 




l 

n 






20 











h 
k 



r 





21 

B 
















t 


z 
22 





H 







l 

n 






23 




E 













t 


z 
24 
A 





a 









r 




z 
\ 
0 
0 
0 
D 
0 
0 
0 
0 
0 
d 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
z 
The conclusion in bold
type after line 24 is –D + d
+ z, which is –D +
(dz), or D É (dz), that is,
“No D is (dz).” Carroll’s conclusion is
transposed and in positive form, “All dz are –D.” This, of course, is just –D + (dz)
alternated to (dz)
+ D, or (dz) É D,
“All dz are –D.” This latter expression
can be changed by obversion to “No dz are D.”
Since the universal negative allows for simple conversion, our
conclusion, “No D is (dz)” is equivalent to Carroll’s, “All dz are –D.”
13. Enthymemes
It is fairly easy to
use logical calculus to find an implicit premiss. The following example is from David Kelley’s The Art of
Reasoning, p. 241:
1. Premiss 1: [ ]
2. Premiss 2: Jane’s
car is a car with a V8 engine
3. Conclusion: Jane’s
car gets poor gas mileage.
The dictionary is: C =
Jane’s car; V=Car with V8 engine; P=poor gas mileage. If we convert the above argument into
symbols, we have:
1. Premiss 1: [ ]
2. Premiss 2: C
+ V
3. Conclusion: C + P
By looking at the
conclusion, we can see that the C
of the conclusion is already in premiss 2, so it doesn’t need to be added to
premiss 1. We can therefore see that V
and P are the two symbols that need to be added to the first premiss. The + P in the conclusion is almost
selfexplanatory: just put it in the first premiss so that it lines up with the
+ P of the conclusion. The V is not in
the conclusion so it was cancelled.
This obviously means it had a polarized form in the first premiss. Hence, the correct missing premiss is:
1. Premiss 1: V + P
2. Premiss 2: C
+ V
3. Conclusion: C + P
Premiss 1, V + P, or All V
is P, says that “All cases of cars with V8 engines are cases of things that
have poor gas mileage.” This is the
equivalent of Kelly’s premiss that says, “Any car with a V8 engine gets poor
gas mileage.”
A logic grid might make
the missing premiss easier to construct:

C 
P 
V 
1 



2 
C 
V 

\ 
C 

P 
Obviously, for the grid to work, a V and a P must be placed in the first line of the grid, as follows:

C 
P 
V 
1 

V 
P 
2 
C 
V 

\ 
C 

P 
The first premiss, of
course, should be read as V
+ P, or All V is P.
The End.