Logical Calculus, Part 3

 

By Vern Crisler

Copyright 2000; 2005; slightly modified, 2008

Rough Draft

 

1.  Simplification

2.  Invalidity

3.  Logical Calculus & Logical Algebra

4.  Solving the Sorites in Logical Calculus

5.  Enthymemes

 

 

1.  Simplification

 

If we want to simplify the following, how would we do it?  Besides the rules in the previous section, there are also 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 + (b-b) = a

 

1*

 

 

2.  (ab) + (a-b) = 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) = a-a + 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

a-b-c

-abc

-a-bc

 

Simply as follows:

 

             1                            2                              3                         4                       5

 

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

 

 

Note: In the second series just above, under premiss 3, the “numerator” should include a “c” but I’ve had problems with html, and it may not show up on some browsers.  It should be the same as premiss 3 in the first series.

 

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.”

 

 

2.  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.

 

AAO-1

 

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.

 

AIA-1

 

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

 

 

 

AOA-1

 

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

 

 

EEO-1

 

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

 

 

 

AAA-2

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: 

 

OAI-2   

 

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:

 

AEO-1

 

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 OAI-2 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 OAI-2 is still invalid, since its conclusion S & P is not equivalent to the correctly derived ~S & P.)

 

So also with the second example, AEO-1.  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, AEO-1 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.

 

4.  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 bold-italic 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 place-holders 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 below, 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.”  Theoretically, all the letters can be represented on two rows (representing plus and minus letters) and it may be convenient to reduce the grid as follows:

 

	A	B	C	D	E	F	G	H	K	L
1	-A	B	C	D	E	-F	G	-H	-K	L
2		-B	-C	-D	-E	F	-G	H	K
	-A	0	0	0	0	0	0	0	0	L

 

In other words, if you already have (say) the letter B in a slot, there is no need to repeat it on other rows, even though it might be in another premiss.  Same with -D or E or -H, and so on,  You only need one positive and one negative to eliminate a term, so repeating the same terms in multiple rows isn’t necessary.  Once you’ve got a positive and a negative letter, you know the terms are polarized and do not come down to the conclusion line.  We have used larger grids only to make it easier to follow the several premises, but these are often inconvenient and it’s easier to use the two-row grid when solving large soriteses.

 

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
Cl1E´0 †	Av´1D0 †		k1m´0 †		lC´1(b´n´)´0 †		dsb1t´0 †	tD1w´0 †
7	8		9			10	11	12
dr´a´1A´0 †	vw1B0 †		em´1(r´b´)´0 †			Ha1c´0 †	dtmav´0 †	dst1A´0 †
13		14		15		16	17	18
Dn´r´b´1z0 †		cE´z0 †		bs´1l´e´0 †		atE1v´0 †	rDh´1e´0 †	mt´1D0 †
19	20			21		22	23	24
Anl´1c´0  †	rdk´1h0 †			ztB´1d0 †		nl´1H´0 †	Et´1z0 †	dzrA´1a´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.  (Because of this I’ve even found some mistakes in Carroll’s logic problems that Bartley overlooked.)

 

 

	Hypothetical	Disjunctive	Neo-Boolean
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 Neo-Boolean 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.”

 

Note: You can see why using a two row grid would make it a lot easier to represent the above problem.  Twenty four rows is way too much information, and a simple positive and negative letter grid would work just as well, including eliminating the 0’s in the conclusion line:

 

A

B

C

D

E

H

a

b

c

d

e

h

k

l

m

n

r

s

t

v

w

z

1

A

-B

-C

-D

E

-H

a

-b

c

-d

-e

h

-k

-l

m

-n

r

-s

t

v

w

-z

2

-A

B

C

-E

H

-a

b

-c

e

-h

k

l

-m

n

-r

s

-t

-v

-w

\

-D

-d

-z

 

I usually keep the order of the negative and positive letters that are found in the premises.  It helps in those cases where I’ve made a mistake and want to retrace my steps.  However, it’s just as easy to put all the positives on one row and all the negatives on the other, as in:

 

+

A

B

C

E

H

a

b

c

e

h

k

l

m

n

r

s

t

v

w

-

-A

-B

-C

-D

-E

-H

-a

-b

-c

-d

-e

-h

-k

-l

-m

-n

-r

-s

-t

-v

-w

-z

\

-D

-d

-z

 

 

Using the two-line grid has enabled me to find a number of (careless) mistakes in Carroll’s Symbolic Logic (and even some mistakes made by his editor Bartley.

 

 

5.  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 V-8 engine

3.  Conclusion:            Jane’s car gets poor gas mileage.

 

The dictionary is: C = Jane’s car; V=Car with V-8 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 self-explanatory: 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 V-8 engines are cases of things that have poor gas mileage.”  This is the equivalent of Kelly’s premiss that says, “Any car with a V-8 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.