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 + (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
_{}
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.
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.
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 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 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 tworow 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 

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 obviousľis 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 
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.”
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 twoline 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 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.