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Logical Calculus, Part 2


by Vern Crisler


Copyright 2000; 2005


6.  Multi-termed 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:

































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´10 † ka´10 † le1m0 † dh10 † h´la´0 † hm´10 † a´bn0 † am´1e0


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:


1st premiss:  ~p v ~q

2nd 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.,







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 (j-j + j-k + -jk + k-k), which can be reduced to j-k + -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 -p-q.  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 -p-q 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 (-p-p + -p-q + -pq + q-q), 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 that--conjoined 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 fell-swoopers, 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





1.  ~p + q

2.  +p      

3.   0 +  q



2)  Modus Tollens


p É q





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






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






8)  Conjunction


p;  q

\p · q


p; q



9)  Addition



\p v q




8.  Logically equivalent or bi-univocal 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.):





~(B v C)

/\ ~A



The conclusion ~A~C is reduced to ~A by simplification.


1.  ~A        + B

2.      ~C & ~B

3.  ~A &~C+0





~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





(G É H) · (I É H)

J É (G v I)


/\ 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




(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 vis-à-vis KL, and can be cancelled.  Also, the conclusion M · N can be simplified to M.


1.      KL + MN

2.  ~(KL)            

3.     0     + MN





(O É P) · (Q É R)

(P É S) · (R É T)

S É ~T


/\ ~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






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.




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.






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








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:






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