510 Taking into account the result arrived at above with regard to A, it will be seen that this may be resolved into Whatever is bC or bD is A and Nothing is BCD or Bcd. These two propositions taken together with the solution for A are equivalent to the original premisses.
489. The Problem of Elimination.—By elimination in logic is meant the omission of certain elements from a proposition or set of 509 propositions with the object of expressing more directly and concisely the connexion between the elements which remain. An example of the process is afforded by the ordinary categorical syllogism, where the so-called middle term is eliminated. Thus, given the premisses All M is P, All S is M, we may infer All S is MP ; but if we desire to know the relation between S and P independently of M we are content with the less precise but sufficient statement All S is P ; in other words, we eliminate M.
Elimination has been considered by some writers to be absolutely essential to logical reasoning. It is not, however, necessarily involved either in the process of contraposition or in the process discussed in the preceding section; and if formal inferences are recognised at all, the name of inference certainly cannot be denied to these processes. We must, therefore, refuse to regard elimination as of the essence of reasoning, although it may usually be involved therein.511
490. Elimination from Universal Affirmations.—Any universal affirmative proposition (or, by combination, any set of universal affirmative propositions) involving the term X and its contradictory x may by contraposition be reduced to the form Everything is PX or Qx or R, where P, Q, R are themselves simple or complex terms not involving X or x ; and since by the rule given in section 448 a determinant may at any time be omitted from an undistributed term, we may eliminate X (and x) from this proposition by simply omitting them, and reducing the proposition to the form Everything is P or Q or R.512
512 We might also proceed as follows: Solve for X and for x, as in section 488, so that we have All X is A, All x is B, where A and B are simple or complex terms not involving either X or x. Then, since Everything is X or x, we shall have Everything is A or B, and this will be a proposition containing neither X nor x.
We must, however, here admit the possibility of P, Q, R being of the forms A or a, Aa. These are equivalent respectively to the entire universe of discourse and to nothing. Thus, if P is of the form A or a, and Q is of the form Aa, our proposition will before elimination more naturally be written Everything is X or R ; if Q is of the form A or a, and R of the form Aa, it will more naturally be written Everything is PX or x. It follows that if either P or Q is of the form A or a (that is, if either P or Q is equivalent to the entire universe of discourse), the proposition resulting from elimination 510 will not afford any real information, since it is always true à priori that Everything is A or a or &c. Thus we are unable to eliminate X from such a proposition as All A is X or BC.
The following may be given as an example of elimination from universal affirmatives.
Let it be required to eliminate X (together with x) from the propositions All P is XQ or xR, Whatever is X or R is p or XQR. Combining these propositions, we have Everything is XQR or p ; therefore, by elimination, Everything is QR or p that is, All P is QR. It will be observed that P (together with p) cannot be eliminated from the above propositions.
491. Elimination from Universal Negatives.—Any universal negative proposition (or, by combination, any set of universal negative propositions) containing the term X and its contradictory x may by conversion be reduced to the form Nothing is PX or Qx or R, where P, Q, R are themselves simple or complex terms not involving either X or x. Here we might, in accordance with the rule given in section 448, simply omit the alternants PX, Qx, leaving us with the proposition Nothing is R. This, however, is but part of the information obtainable by the elimination of X. We have also No X is P, and No Q is x, that is, All Q is X ; whence by a syllogism in Celarent we may infer No Q is P. The full result of the elimination is, therefore, given by the proposition Nothing is PQ or R.513
513 Compare Mrs Ladd Franklin’s Essay on The Algebra of Logic (Studies in Logic by Members of the Johns Hopkins University). The same conclusion may be deduced by obversion from the result obtained in the preceding section. Nothing is PX or Qx or R becomes by obversion Everything is prX or qrx. Therefore, by the elimination of X, Everything is pr or qr ; and this proposition becomes by obversion Nothing is PQ or R.
Another method by which the same result may be obtained is as follows: By developing the first alternant with reference to Q and the second with reference to P, Nothing is PX or Qx or R becomes Nothing is PQX or PqX or PQx or pQx or R. But PQX or PQx is reducible to PQ, and on omitting PqX and pQx, we have Nothing is PQ or R.
It is interesting to observe that the above rule for elimination from negatives is equivalent to Boole’s famous rule for elimination. In order to eliminate X from the equation F(X) = 0, he gives the formula F(1) F(0) = 0. Now any equation containing X can be brought to the form AX + Bx + C = 0, where A, B, C are independent of X. Applying Boole’s rule we have (A + C)(B + C) = 0, that is, AB + C = 0; and this is precisely equivalent to the rule given in the text.
The following is an example: Let it be required to eliminate X from the propositions No P is Xq or xr, No X or R is xP or Pq or Pr. 511 Combining these propositions we have Nothing is XPq or XPr or xP or PqR ; therefore, by elimination in accordance with the above rule, Nothing is Pq or Pr, that is, No P is q or r.
492. Elimination from Particular Affirmatives.—Any particular affirmative proposition involving the term X may by conversion be reduced to the form Something is either PX or Qx or R, where P, Q, R are independent of X and x. We may here immediately apply the rule given in section 448 that a determinant may at any time be omitted from an undistributed term; and the result of eliminating X is accordingly Something is either P or Q or R.514
514 Thus the rule for elimination from particular affirmatives is practically identical with the rule for elimination from universal affirmatives.
493. Elimination from Particular Negatives.—Any particular negative proposition involving the term may by contraposition be reduced to the form Something is not either PX or Qx or R. By the development of the first alternant with reference to Q and that of the second alternant with reference to P, this proposition becomes Something is not either PQX or PqX or PQx or pQx or R. But PQX or PQx is reducible to PQ and the alternants PqX, pQx may by the rule given in section 448 be omitted. Hence we get the proposition Something is not either PQ or R, from which X has been eliminated.515
515 Thus the rule for elimination from particular negatives is practically identical with the rule for elimination from universal negatives. The same rule may be deduced by obversion from the result obtained in the preceding section. Something is not either PX or Qx or R ; therefore, Something is either prX or qrx or pqr ; therefore, Something is either pr or qr ; therefore, Something is not either PQ or R.
494. Order of procedure in the process of elimination.—Schröder (Der Operationskreis des Logikkalkuls, p. 23) points out that first to eliminate and then combine is not the same thing as first to combine and then eliminate. For, as a rule, if a term X is eliminated from several isolated propositions the combined results give less information than is afforded by first combining the given propositions and then effecting the required elimination.
There are indeed many cases in which we cannot eliminate at all unless we first combine the given propositions. This is of course obvious in syllogisms; and we have a similar case if we take the premisses Everything is A or X, Everything is B or x. We cannot eliminate X from either of these propositions taken by itself, since in each of them X (or x) appears as an isolated alternant. But by 512 combination we have Everything is Ax or BX ; and this by the elimination of X becomes Everything is A or B.516
516 Working with negatives we get the same result. Taking the propositions Nothing is ax, Nothing is bX, separately, we cannot eliminate X from either of them. But combining them in the proposition Nothing is ax or bX, we are able to infer Nothing is ab.
There are other cases in which elimination from the separate propositions is possible, but where this order of procedure leads to a weakened conclusion. Take the propositions Everything is AX or Bx, Everything is CX or Dx. By first eliminating X and then combining, we have Everything is AC or AD or BC or BD. But by first combining and then eliminating X our conclusion becomes Everything is AC or BD, which gives more information than is afforded by the previous conclusion.
EXERCISES.
495. Suppose that an
analysis of the properties of a particular class of substances has led
to the following general conclusions, namely:
1st, That wherever the properties A and B are
combined, either the property C, or the property D, is
present also; but they are not jointly present;
2nd, That wherever the properties B and C are
combined, the properties A and D are either both present
with them, or both absent;
3rd, That wherever the properties A and B are both
absent, the properties C and D are both absent also; and
vice versâ, where the properties C and D are both
absent, A and B are both absent also.
Shew that wherever the property A is present, the properties
B and C are not both present; also that wherever
B is absent while C is present, A is present.
[Boole, Laws of Thought, pp. 118 to 120; compare also Venn, Symbolic Logic, pp. 276 to 278.]
A solution of this problem has already been given in section 488. We may also proceed as follows. The premisses are:
All AB is Cd or cD, (i)
All BC is AD or ad,(ii)
All ab is cd,(iii)
All cd is ab.(iv)
513 By (i), No AB is CD, therefore, No A is BCD. (1)
By (ii), No BC is Ad, therefore, No A is BCd. (2)
Combining (1) and (2), it follows immediately that No A is BC.
Boole also shews that All bC is A. This is a partial contrapositive of (iii). We have so far not required to make use of (iv) at all.
496. Given the same
premisses as in the preceding section, prove that:—
(1) Wherever the property C is found, either the property
A or the property B will be found with it, but not both
of them together;
(2) If the property B is absent, either A and
C will be jointly present, or C will be absent;
(3) If A and C are jointly present, B will be
absent.[Boole, Laws of Thought, p. 129.]
First, By (i), All C is a or b or d ; by (ii), All C is a or b or D ; therefore, All C is a or b.
Also, by (iii), All C is A or B ;
therefore, All C is Ab or aB. (1)
Secondly, By (iii). All b is A or c,
therefore, by section 432, All b is AC or c. (2)
Thirdly, from (1) it follows immediately that
All AC is b. (3)
The given premisses may all be summed up in the proposition: Everything is AbC or AbD or aBCd or abcd or BcD. From this, the above special results and others follow immediately.
497. Given that everything is either Q or R, and that all R is Q, unless it is not P, prove that all P is Q. [K.]
The premisses may be written as follows: (1) All r is Q, (2) All PR is Q.
By (1), All Pr is Q, and by (2), All PR is Q ; but
All P is Pr or PR ; therefore, All P is Q.
498. Where A is present, B and C are either both present at once or absent at once; and where C is present, A is present. Describe the class not-B under these conditions. [Jevons, Studies, p. 204.]
The premisses are (1) All A is BC or bc, (2) All C is A.
By (1) All b is a or c, and by (2) All b is A or c,
therefore, All b is c.
499. It is known of certain things that (1) where the quality A is, B is not; (2) where B is, and only where B is, C and D are. 514 Derive from these conditions a description of the class of things in which A is not present, but C is. [Jevons, Studies, p. 200.]
The premisses are: (1) All A is b ; (2) All B is CD ; (3) All CD is B.
No information regarding aC is given by
(1). But by (2), All aC is b or D ; and by (3), All aC is B or d.
Therefore, All aC is BD or bd.
500. Taking the same premisses as in the previous section, draw descriptions of the classes Ac, ab, and cD. [Jevons, Studies, p. 244.]
By (1), Everything is a or b, and by (2), Everything is b
or CD. Therefore, Everything is aCD or b ; and by (3), Everything is B or c or d. Therefore, Everything is aBCD or bc or bd.
Hence we infer immediately All Ac is b, All ab
is c or d, All cD is b.
501. There is a certain class of things from which A picks out the ‘X that is E, and the Y that is not Z,’ and B picks out from the remainder ‘the Z which is Y and the X that is not Y.’ It is then found that nothing is left but the class ‘Z which is not X.’ The whole of this class is however left. What can be determined about the class originally? [Venn, Symbolic Logic, pp. 267, 8.]
The chief difficulty in this problem consists in the accurate statement of the premisses. Call the original class W. We then have
All W is XZ or Yz or YZ or Xy or xZ,
that is, All W is X or Y or Z ; (1)
All xZ is W ; (2)
No xZ is WXZ or WYz or WYZ or WXy,
that is, No xZ is WYZ. (3)
We may now proceed as follows:—By (1), All W is X or Y or Z ; and by (3), All W is X or y or z. Therefore, All W is X or Yz or yZ. (2) affords no information regarding the class W, except that everything that is Z but not X is contained within it.
502. (1) If a nation has
natural resources, and a good government, it will be prosperous. (2)
If it has natural resources without a good government, or a good
government without natural resources, it will be contented, but not
prosperous. (3) If it has neither natural resources nor a good
government it will be neither contented nor prosperous.
Shew that these statements may be reduced to two propositions of
the form of Hamilton’s U. [O’S]
515 Let a nation with natural resources be denoted by R, a
nation with a good government by G, a prosperous nation by
P, and a contented nation by C. Then the given
statements may be expressed as follows:—(1) All RG is P ;
(2) All Rg or rG is Cp ; (3) All rg is cp.
By contraposition, (2) may be resolved into the two propositions, All
cp is RG or rg, All P is RG or rg. But by (1) No cp is
RG ; and by (3) No P is rg. Hence the two propositions into
which (2) was resolved may be reduced to the form, All cp is
rg, All P is RG.
The three original statements are accordingly equivalent to the two U propositions All RG is all P, All rg is all cp.
503. Let the observation of
a class of natural productions be supposed to have led to the
following general results.
1st. That in whichsoever of these productions the properties
A and C are missing, the property E is found,
together with one of the properties B and D, but not
with both.
2nd. That wherever the properties A and D are found
while E is missing, the properties B and C will
either both be found, or both be missing.
3rd. That wherever the property A is found in conjunction
with either B or E, or both of them, there either the
property C or the property D will be found, but not both
of them. And conversely, wherever the property C or D is
found singly, there the property A will be found in conjunction
with either B or E or both of them.
Shew that it follows that In whatever substances the property A
is found, there will also be found either the property C or the
property D, but not both, or else the properties B, C, and E will all
be wanting. And conversely, Where either the property C or the
property D is found singly or the properties B. C, and D are together
missing, there the property A will be found. Shew also that If
the property A is absent and C present, D is present.
[Boole, Laws of Thought, pp. 146–148. Venn, Symbolic Logic, pp. 280, 281. Johns Hopkins Studies in Logic, pp. 57, 58, 82, 83.]
The premisses are as follows:—
| 1st, | All ac is BdE or bDE ; | (i) |
| 2nd, | All Ade is BC or bc ; | (ii) |
| 3rd, | Whatever is AB or AE is Cd or cD ; | (iii) |
| Whatever is Cd or cD is AB or AE. | (iv) |
516 We are required to prove:—
| All A is Cd or cD or bcd ; | (α) |
| All Cd is A ; | (β) |
| All cD is A ; | (γ) |
| All bcd is A ; | (δ) |
| All aC is D. | (ε) |
First, By (iii), All A is Cd or cD or bc. But by (ii), All Abe is c or d ; and by (iv), All Abe is CD or cd ; therefore, All Abe is cd. Hence, All A is Cd or cD or bcd. (α)
Secondly, (β) and (γ) follow immediately from (iv).
Thirdly, from (i), we have directly, No ac is bd ; therefore (by
conversion), No bcd is a ; therefore, All bcd is A.
(δ)
Lastly, by (iv), All Cd is A ; therefore, by contraposition, All aC is D. (ε)
We may obtain a complete solution so far as A is concerned as follows:
By (ii),517 All A is BC or
bc or d or E ;
by (iii), All is be or Cd or cD ;
therefore,
All A is Cd or cDE or bcD or bce or bde ;
by (iv). All A is B
or E or CD or cd ;
therefore, All A is cDE or bcde or BCd or
CdE.
This includes the partial solution with regard to
A,—All A is Cd or cD or bcd. Boole contents
himself with this because he has started with the intention of
eliminating E from his conclusion.
We may now solve for
a. (ii) and (iii) give no information with regard to this term.
But by (i), All a is BdE or bDE or C ; and by (iv), All a is
CD or cd. Therefore, All a is BcdE or CD. And this yields
by contraposition, Whatever is bc or Cd or cD or ce is A.
517 No information whatever with regard to A is given by (i), since a appears as a determinant of the subject. See section 487.
504. Given the same
premisses as in the preceding section, shew that,—
1st. If the property B be present in one of the productions,
either the properties A, C, and D are all absent, or some one alone of
them is absent. And conversely, if they are all absent it may be
concluded that the property B is present.
2nd. If A and C are both present or both absent, D will be
absent, quite independently of the presence or absence of B.
[Boole, Laws of Thought, p. 149.]
We may proceed here by combining all the given premisses in 517 the manner indicated in section 475. From the result thus obtained the above conclusions as well as those contained in the preceding section will immediately follow.
By (iii), Everything is a or be or Cd or cD ;
and by (iv).
Everything is AB or AE or CD or cd ;
therefore, Everything is ABCd or ABcD or ACdE or AcDE or aCD or acd or bCDe or bcde ;
therefore by (i), Everything is ABCd or ABcD or Abcde or ACdE or AcDE or aBcdE or aCD or bCDe ;
therefore by (ii), Everything is ABCd or Abcde or ACdE or AcDE or aBcdE or aCD.
(v)
Hence, All B is ACd or AcDE or acdE or aCD ;
All acd is BE ;
All AC is Bd or dE ;
All ac is BdE.
Eliminating E from each of the above we have the results arrived at by Boole.
Eliminating both A and E from (v) we have
Everything is BCd or bcd or Cd or cD or Bcd or CD ;
that is Everything is C or D or cd, which is an identity. This is equivalent to Boole’s conclusion that “there is no
independent relation among the properties B, C, and
D” (Laws of Thought, p. 148).
Any further results that may be desired are obtainable immediately
from (v).
505. Given XY = A, YZ = C, find XZ in terms of A and C.
[Venn, Symbolic Logic, pp. 279, 310–312. Johns Hopkins Studies in Logic, pp. 53, 54.]
The premisses may be written as follows:
Everything is AXY or ax or ay ; (1)
Everything is CYZ or cy or cx. (2)
By (1), All XZ is AY or ay, and by (2), All XZ
is CY or cy ; therefore, All XZ is ACY or acy. Hence,
eliminating Y, All XZ is AC or ac.
This solves the problem as set. But in order to get a complete
solution equivalent to that which would be obtained by Boole, the
following may be added: Solving as above for x or z, and
eliminating Y, we have All that is either x or z is AcXz or
aCxZ or ac. Whence, by contraposition, Whatever is AC or Ax or
AZ or CX or Cz is XZ. In other words, Whatever is AC or AZ or
CX is XZ ; and Nothing is Ax or Cz.
518 506. Shew the equivalence between the three following systems of propositions: (1) All Ab is cd ; All aB is Ce ; All D is E ; (2) All A is B or c or D ; All BE is A ; All Be is Ad or Cd ; All bD is aE ; (3) Whatever is A or e is B or d ; All a is bE or bd or BCe ; All bC is a ; All D is E. [K.]
By obversion, the first set of propositions become No Ab is C or D ; No aB is c or E ; No D is e ; and these propositions are combined in the statement, Nothing is either AbC or AbD or aBc or aBE or De. (1)
By obverting and combining the second set of propositions, we have Nothing is AbCd or aBE or aBce or BDe or AbD or bDe. (2)
But AbCd or AbD is equivalent to
AbC or AbD ; aBE or aBce to aBE or aBc ; BDe
or bDe to De. Hence (1) and (2) are equivalent.
Again, by obverting and combining the third set of propositions, we have
Nothing is AbD or bDe or aBc or aBE or abDe or acDe or AbC or
De. (3)
But since bDe, abDe, acDe are all subdivisions of De, (3) immediately resolves itself into (1).
507. From the premisses (1) No Ax is cd or cy, (2) No BX is cde or cey, (3) No ab is cdx or cEx, (4) No A or B or C is xy, deduce a proposition containing neither X nor Y. [Johns Hopkins Studies, p. 53.]
By (2), No X is Bcde, and by (1) and (3), No x is Acd or abcd or abcE ; therefore, by section 491, No Acd or abcd or abcE
is Bcde ; therefore, No Acd is Be.
It will be observed that since Y does not appear in the
premisses, y can be eliminated only by omitting all the terms
containing it.
508. The members of a scientific society are divided into three sections, which are denoted by A, B, C. Every member must join one, at least, of these sections, subject to the following conditions: (1) any one who is a member of A but not of B, of B but not of C, or of C but not of A, may deliver a lecture to the members if he has paid his subscription, but otherwise not; (2) one who is a member of A but not of C, of C but not of A, or of B but not of A, may exhibit an experiment to the members if he has paid his subscription, but otherwise not; but (3) every member must either deliver a lecture or perform an experiment annually before the other members. Find the least addition to these rules which will compel every member to pay his subscription or forfeit his membership. [Johns Hopkins Studies, p. 54.]
Let A = member of section A, &c.; X = one who gives a lecture; 519 Y = one who performs an experiment;
Z = one who has paid his subscription.
The premisses are
(1) All Ab or aC or Bc is x or Z ;
(2) All Ac or aB or aC is y or Z ;
(3) Every member is X or Y ;
(4) Every member is A or B or C.
The problem is to find what is the least addition to these
rules which will result in the conclusion that Every member is
Z.
By (1), All z is either x or else (a or B) (A or
c) (b or C);
therefore, All z is x or ABC or abc.
Similarly, by (2), All z is y or AC or abc ;
therefore, All z is xy or xAC or ABC or abc.
By (3), All z is X or Y ;
therefore,All z is XABC or Xabc or xYAC or YABC or Yabc.
By (4), All z is A or B or C ;
therefore, All z is XABC or xYAC
or YABC ;
but All YABC is either XYABC or xYABC ;
therefore, All z is XABC or xYAC.
Hence, we gain the desired result if we
add to the premisses, No z is XABC or xYAC. The required rule
is therefore as follows: No one who has not paid his subscription
may join all three sections and deliver a lecture, nor may he join A
and C and exhibit an experiment without delivering a lecture.
509. What may be inferred independently of X and Y from the premisses: (1) Either some A that is X is not Y, or all D is both X and Y ; (2) Either some Y is both B and X, or all X is either not Y or C and not B? [Johns Hopkins Studies, p. 85.]
The premisses may be written as follows: (1) Either something is
AXy, or everything is XY or d ; (2) Either something is BXY, or
everything is x or y or bC.
By combining these premisses as in
chapter 4, Either something is AXy and something is BXY, or
something is AXy and everything is x or y or bC, or something is BXY
and everything is XY or d, or everything is bCXY or bCd or dx or
dy.518
Therefore, eliminating X and Y
(see sections 490 and 492), Either something is A and something is
B, or something is A, or 520 something is B, or everything is bC or
d ; and by combining the first three alternants as in section 481, this becomes
Either something is A or B or everything is bC or d.
This conclusion may also be expressed in the form
If everything is ab, then every c is d.
518 We cannot, if we are to be left with an equivalent proposition, express the first three of these alternants in a non-compound form. See sections 477, 479.
510. Six children, A, B, C, D, E, F are required to obey the following rules: (1) on Monday and Tuesday no four can go out together; (2) on Thursday, Friday, and Saturday no three can stay in together; (3) on Tuesday, Wednesday, and Saturday, if B and C are together, then A, B, E, and F must be together; (4) on Monday and Saturday B cannot go out unless either D, or A, C, and E stay at home. A and B are first to decide what they will do, and C makes his decision before D, E, and F. Find (α) when C must go out, (β) when he must stay in, and (γ) when he may do as he pleases. [Johns Hopkins Studies, p. 58.]
Let A = case in which A goes out, a = that in which he stays in, &c.
Then the premisses are as follows:
(1) On Monday and Tuesday,—three at least must stay in ;
(2) On Thursday, Friday, and Saturday,—no three can stay in
together ;
(3) On Tuesday, Wednesday, and Saturday,—Every
case is ABEF or abef or Bc or bC ;
(4) On Monday and Saturday,—Every case is ace or b or d.
In order to solve the problem, we must combine the possibilities
for each day, then eliminate D, E, and F, and
find in what ways the movements of A and B determine
those of C.
(i) On Monday,—we have Every case is ace or b or d,
combined with the condition that three at least must stay in. One
alternant therefore is def without further condition, and it
follows that we can determine no independent relation between
A, B, and C.
Hence on Monday C may do as he pleases.
(ii) On Tuesday,—we have Every case is ABEF or abef or Bc
or bC, combined with the condition that three at least must stay
in. Therefore, Every case is abef or Bc or bC ;519
and eliminating D, E, and F, Every case is ab
or Bc or bC.
519 The two alternants Bc and bC might here be made more determinate, thus, aBcd or aBce or aBcf or Bcde or Bcdf or Bcef and abCd or abCe or abCf or bCde or bCdf or bCef. But since we know that we are going on immediately to eliminate d, e, and f, it is obvious, even without writing them out in full, that these more determinate expressions will at once be reduced again to Bc and bC simply.
521 Hence it
follows that on Tuesday (α) if A goes out while B
stays in, C must go out, and (β) if B goes out, C must
stay in.
(iii) On Wednesday,—Every case is ABEF or abef or Bc or
bC ; or, eliminating D, E, and F, Every
case is AB or ab or Bc or bC. Therefore, All Ab is C and
All aB is c.
Hence on Wednesday (α) if A
goes out while B stays in, C must go out, and (β) if A
stays in while B goes out, C must stay in.
(iv) On Thursday and Friday,—the only condition is that no
three can stay in together.
Hence on Thursday and Friday if A and B both stay in, C must go out.
(v) On Saturday,—Every case is ABEF or abef or Bc or
bC ; also Every case is ace or b or d. Combining these
premisses, Every case is ABdEF or abef or aBce or Bcd or bC.
But we have the further condition that no three can stay in together.
Therefore, Every case is ABdEF or ABcdEF or AbCDE or AbCDF or AbCEF
or bCDEF. Therefore, eliminating D, E, and F,
Every case is AB or bC.
Hence on Saturday if B stays in, C must go out.
511. Given (1) All P is QR, (2) All p is qr ; shew that (3) All Q is PR, (4) All R is PQ. [K.]
512. Eliminate R from the propositions All R is P or pq, All q is Pr or R, All qR is P. [K.]
513. Shew the equivalence between the following sets of propositions:—(1) a is BC ; b is AC ; C is Ab or aB ; (2) a is BC ; B is Ac or aC ; c is AB ; (3) A is Bc or bC ; b is AC ; c is AB. [K.]
514. Say by inspection, stating your reasons, which of the following propositions give information concerning A, aB, b, bCd, respectively: All Ab is bCd or c ; All bd is A or bC or abc ; Whatever is a or B is c or D ; Whatever is Ab or bc is bD or cD or e ; Everything is A or ab or Bc or Cd. [K.]
515. Determine the conditions under which a particular proposition affords information in regard to any given term. [K.]
516. It is known of certain things that the quality A is always accompanied by C and D, but never by B ; and further, that the qualities C and D never occur together, except in conjunction with A. What can we infer about C? [M.]
522 517. Given that everything that is Q but not S is either both P and or neither P nor R and that neither R nor S is both P and Q, shew that no P is Q. [K.]
518. Where C is present, A, B, and D are all present; where D is present, A, B, and C are either all three present or all three absent. Shew that when either A or B is present, C and D are either both present or both absent. How much of the given information is superfluous so far as the desired conclusion is concerned? [K.]
519. Given (i) All Pqr is ST ; (ii) Q and R are always present or absent together ; (iii) All QRS is PT or pt ; (iv) All QRs is Pt ; (v) All pqrS is T ; then it follows that (1) All Pq is rST ; (2) All Ps is QRt ; (3) All pQ is RSt ; (4) All pT is qr ; (5) All Qs is PRt ; (6) All QT is PRS ; (7) All qS is rT ; (8) All qs is pr ; (9) All qt is prs ; (10) All sT is pqr. [K.]
520. What can be determined about P in terms of Q and R from the premisses All P is Q or X, Some P is not RX? [K.]
521. Given that all honest men are happy, and that all dishonest men are unwise; and assuming that honest and dishonest, happy and unhappy, wise and unwise, are pairs of logical contradictories; what is all that can be inferred about men who are happy, unhappy, wise, unwise, respectively? [K.]
522. If thriftlessness and poverty are inseparable, and virtue and misery are incompatible, and if thrift be a virtue, can any relation be proved to exist between misery and poverty? If moreover all thriftless people are either virtuous or not miserable, what follows? [V.]
523. At a certain examination, all the candidates who were entered for Latin were also entered for either Greek, French, or German, but not for more than one of these languages; all the candidates who were not entered for German were entered for two at least of the other languages; no candidate who was entered for both Greek and French was entered for German, but all candidates who were entered for neither Greek nor French were entered for Latin. Shew that all the candidates were entered for two of the four languages, but none for more than two. [K.]
524. (1) Wherever there is smoke there is also fire or light; (2) Wherever there is light and smoke there is also fire; (3) There is no fire without either smoke or light.
523 Given the truth of the above propositions, what is all that you can infer with regard to (i) circumstances where there is smoke; (ii) circumstances where there is not smoke; (iii) circumstances where there is not light? [W.]
525. In a certain warehouse, when the articles offered are antique, they are costly, and at the same time either beautiful or grotesque, but not both. When they are both modern and grotesque, they are neither beautiful nor costly. Everything which is not beautiful is offered at a low price, and nothing cheap is beautiful. What can we assert (1) about the antique, and (2) about the grotesque articles? [M.]
526. Shew that the following
sets of propositions are equivalent to one another:—
(1) All a is b or c ; All b is aCd ; All c is aB ; All D is c.
(2) All A is BC ; All b is aC ; All C is ABd or abd.
(3) All A is B ; All B is A or c ; All c is aB ; All D is c.
(4) All b is aC ; All A is C ; All C is d ;
All aC is b.
(5) All c is aB ; All D is aB ; All A is B ; All aB is c.
(6) All A is BC ; All BC is A ; All D is Bc ; All
b is C. [K.]
527. Shew that a certain set of four properties must be found somewhere together, if the following facts are known: “Everything that has the first property or is without the last has the two others; and if everything that has both the first and last has one or other but not both of the two others, then something that has the first two must be without the last two.” [J.]
528. Given the propositions: (i) all material goods are external; (ii) no internal (= non-external) goods are dispropriable; (iii) all dispropriable goods are appropriable; (iv) no collective goods are appropriable or immaterial (= non-material); what is all that we can infer about (a) appropriable goods, (b) immaterial goods? [J.]
529. Eliminate X and Y from the following propositions: All aX is BcY or bcy ; No AX is BY ; All AB is Y ; No ABCD is xY. Shew also that it follows from these propositions that All XY is Ab or aBc. [K.]
530. Given (1) All A is Bc or bC, (2) All B is DE or de, (3) All C is De ; shew that (i) All A is BcDE or Bcde or bCDe, (ii) All BcD is E, (iii) All abd is c, (iv) All cd is ab or Be, (v) All bCD is e. [Jevons, Pure Logic, § 160.]
524 531. Given (1) All aB is c or D, (2) All BE is DF or cdF, (3) All C is aB or BE or D, (4) All bD is e or F, (5) All bf is a or C or DE, (6) All bcdE is Af or aF, (7) All A is B or CDEf or cDf or cdE ; shew that (i) All A is B, (ii) All C is D, (iii) All E is F. [K.]
532. Shew the equivalence between the two following sets of propositions:
| (1) | All A is BC or BE or CE or D ; |
| All B is ACDE or ACde or cdE ; | |
| All C is AB or AE or aD ; | |
| All D is ABCE or Ace or aC ; | |
| All E is AC or aCB or Bc. | |
| (2) | All a is BcdE or bcde or bD ; |
| All b is a or ce or dE ; | |
| All c is AbDe or abde or BdE ; | |
| All d is abce or BcE or Be or bE ; | |
| All e is ab or bc or d. |
533. Given
| (1) | All bc is DE or Df or hk, |
| (2) | All C is aB or DEFG or BFH, |
| (3) | All Bcd is eL or hk, |
| (4) | All Acf is d, |
| (5) | All k is BC or Cd or Cf or H, |
| (6) | All ABCDEFG is H or K, |
| (7) | All DEFGH is B, |
| (8) | All ABl is f or h, |
| (9) | All ADFKl is H, |
| (10) | All ADEFH is B or C or G or L ; |
shew that All A is L.[K.]
CHAPTER VI.
THE INVERSE PROBLEM.
534. Nature of the Inverse Problem.—By the inverse problem is here meant a certain problem so-called by Jevons. Its nature will be indicated by the following extracts, which are from the Principles of Science and the Studies in Deductive Logic respectively.
“In the Indirect process of Inference we found that from certain propositions we could infallibly determine the combinations of terms agreeing with those premisses. The inductive problem is just the inverse. Having given certain combinations of terms, we need to ascertain the propositions with which they are consistent, and from which they may have proceeded. Now if the reader contemplates the following combinations,—
| ABC | abC | |
| aBC | abc, |
he will probably remember at once that they belong to the premisses A = AB, B = BC. If not, he will require a few trials before he meets with the right answer, and every trial will consist in assuming certain laws and observing whether the deduced results agree with the data. To test the facility with which he can solve this inductive problem, let him casually strike out any of the possible combinations involving three terms, and say what laws the remaining combinations obey. Let him say, for instance, what laws are embodied in the combinations,—
| ABC | aBC | |
| Abc | abC, |
“The difficulty becomes much greater when more terms enter 526 into the combinations. It would be no easy matter to point out the complete conditions fulfilled in the combinations,—
| ACe |
| aBCe |
| aBcdE |
| abCe |
| abcE. |
After some trouble the reader may discover that the principal laws are C = e, and A = Ae ; but he would hardly discover the remaining law, namely that BD = BDe” (Principles of Science, 1st ed., vol. I., p. 144; 2nd ed., p. 125).
“The inverse problem is always tentative, and consists in inventing laws, and trying whether their results agree with those before us” (Studies in Deductive Logic, p. 252).
The problem may preferably be stated as follows:—
Given a complex proposition of the form
Everything is P1P2 … or Q1Q2 … or …,
to find a set of propositions not involving any alternative combination of terms, which shall together be equivalent to it.520