Don't Ask Alice - Part the Rhyme

This is the ninth installment on Dodgson's unsolved sorites - for the previous ones go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Don't ask Alice - Part the Fourth [from Dodgson's Symbolic Logic, p. 190]
Don't ask Alice - Part the Fifth [from Dodgson's Symbolic Logic, p. 191]
Don't ask Alice - Part the Sixth [from Dodgson's Symbolic Logic, p. 192]
Don't ask Alice - Part the Seventh [from Dodgson's Symbolic Logic, p. 193]
Don't ask Alice - Part the Eighth [from Dodgson's Symbolic Logic, p. 194
The ninth problem is not one of his compositions, but rather the nursery rhyme:

"Jack Sprat could eat no fat:
His wife could eat no lean:
And so, between them both,
They licked the platter clean."

Solve this as a Sorites-Problem, taking lines 3 and 4 as the Conclusion to be proved. It is permitted to use, as Premisses, not only all that is here asserted, but also all that we may reasonably understand to be implied.

It is difficult to understand what Dodgson had in mind as a sorites proof of this verse. Begin by assuming that the universe is types of food, and define the following classes : J – the food that Jack Sprat eats, W – the food his wife eats, F – foods which are fatty, and L – foods that are lean. It may reasonably be assumed that the classes F and L are complementary, i.e., F = L^~ and vice versa. The premises are interpreted as: firstly, that "None of the food Jack Sprat eats is fatty' or ρ(J^~F^~), and secondly, that 'None of the food his wife eats is lean' or ρ(W^~L^~) = ρ(W^~F). Writing the conjunction of these premises and using the theorem to find the conclusion :
ρ(J^~F^~) · ρ(W^~F) ⇒ ρ(J^~W^~)
But this conclusion is only that Jack ate nothing his wife ate and vice-versa - the two classes are disjoint. The conclusion 'They licked the platter clean' is not valid, since it is possible they did not eat all the food. The purported conclusion follows only on the additional assumptions that Jack ate all the lean and his wife ate all the fat, or that together they ate all the food, in either case begging the question.

Perhaps Dodgson intended this exercise to show that the assumptions did not support the purported conclusion, or perhaps he was simply mistaken - we'll never know.

Don't Ask Alice - Part the Eighth

This is the eighth installment on Dodgson's unsolved sorites - for the previous seven go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Don't ask Alice - Part the Fourth [from Dodgson's Symbolic Logic, p. 190]
Don't ask Alice - Part the Fifth [from Dodgson's Symbolic Logic, p. 191]
Don't ask Alice - Part the Sixth [from Dodgson's Symbolic Logic, p. 192]
Don't ask Alice - Part the Seventh [from Dodgson's Symbolic Logic, p. 193]
Without further ado, here is the eighth of Dodgson's [hitherto] unsolved sorites:

(1) A man can always master his father;
(2) An inferior of a man's uncle owes that man money;
(3) The father of an enemy of a friend of a man owes that man nothing;
(4) A man is always persecuted by his son's creditors;
(5) An inferior of the master of a man's son is senior to that man;
(6) A grandsom of a man's junior is not his nephew;
(7) A servant of an inferior of a friend of a man's enemy is never persecuted by that man;
(8) A friend of a superior of the master of a man's victim is that man's enemy;
(9) An enemy of a persecutor of a servant of a man's father is that man's friend.

The Problem is to deduce some fact about great-grandsons.

[N.B. In this Problem, it is assumed that all the men, here referred to, live in the same town, and that every pair of them are either "friends" or "enemies," that every pair are related as "senior and junior", "superior and inferior", and that certain pairs are related as "creditor and debtor", "father and son", "master and servant", "persecutor and victim", "uncle and nephew".]

Define the following two-place relations: e(X,Y) - X and Y are enemies, assumed symmetric, so that e(X,Y) = e(Y,X), and e(X,Y)^~ means X and Y are friends; s(X,Y) - X is senior to Y, or Y is junior to X, assumed anti-symmetric so s(X,Y) = s(Y,X)^~ and vice-versa; i(X,Y) - X is inferior to Y, or Y is superior to X, assumed anti-symmetric so i(X,Y) = i(Y,X)^~ and vice-versa; c(X,Y) - X is a creditor of Y, Y is a debtor of X, assumed disjoint so c(X,Y)^~c(Y,X)^~ = U; f(X,Y) - X is the father of Y, or Y is the son of X, assumed disjoint so f(X,Y)^~f(Y,X)^~ = U; m(X,Y) - X is a master of Y, or Y is the servant of X, assumed disjoint so m(X,Y)^~m(Y,X)^~ = U; p(X,Y) - X is a persecutor of Y, or Y is a victim of X, assumed disjoint so p(X,Y)^~p(Y,X)^~ = U; and u(X,Y) - X is an uncle of Y, or Y is a nephew of X, assumed disjoint so u(X,Y)^~u(Y,X)^~ = U. We may write the premises as:
1) f(B,A) ⇒ m(A,B) or ρ(f(B,A)^~m(A,B))
2) u(C,B) · i(A,C) ⇒ c(B,A) or ρ(c(B,A)i(A,C)^~u(C,B)^~)
3) e(C,D)^~ · e(B,C) · f(A,B) ⇒ c(D,A)^~ or ρ(c(D,A)^~e(C,D)e(B,C)^~f(A,B)^~)
4) f(A,B) · c(C,B) ⇒ p(C,A) or ρ(c(C,B)^~f(A,B)^~p(C,A))
5) i(A,B) · m(B,D) · f(C,D) ⇒ s(A,C) or ρ(f(C,D)^~i(A,B)^~m(B,D)^~s(A,C))
6) f(C,D) · f(D,A) · s(B,C) ⇒ u(B,A)^~ or ρ(f(C,D)^~f(D,A)^~s(B,C)^~u(B,A)^~)
7) m(B,A) · i(B,C) · e(C,E)^~ · e(D,E) ⇒ p(D,A)^~ or ρ(e(C,E)e(D,E)^~i(B,C)^~m(B,A)^~p(D,A)^~)
8) e(A,B)^~ · i(C,B) · m(C,E) · p(D,E) ⇒ e(A,D) or ρ(e(A,B)e(A,D)i(C,B)^~m(C,E)^~p(D,E)^~)
9) e(A,B) ·p(B,C) · m(E,C) · f(E,D) ⇒ e(A,D)^~ or ρ(e(A,D)^~e(A,B)^~f(E,D)^~m(E,C)^~p(B,C)^~)

It must be kept in mind that the letters used in each premise are mere placeholders, indicating the relationships between the various parameters in the two-place functions but in general different from the same letter in a different premise. The problem is to find substitutions in each premise which will eliminate the most terms, leading to the minimal conclusion. To facilitate this process, construct a table with columns for each function and for each premise show the constituent functions with numbers as the parameter placeholders.

	c(...)	e(...)	f(...)	i(...)	m(...)	p(...)	s(...)	u(...)
1)			f(1,2)~		m(2,1)
2)	c(1,2)			i(2,3)~				u(3,1)~
3)	c(1,2)~	e(3,1) e(4,3)~	f(2,4)~
4)	c(1,2)~		f(3,2)~			p(1,3)
5)			f(1,2)~	i(3,4)~	m(4,2)~		s(3,1)
6)			f(1,2)~ f(2,3)~				s(4,1)~	u(4,3)~
7)		e(1,2) e(3,2)~		i(4,1)~	m(4,5)~	p(3,5)~
8)		e(1,2) e(1,3)		i(4,2)~	m(4,5)~	p(3,5)~
9)		e(1,2)~ e(1,3)~	f(4,3)~		m(4,5)~	p(2,5)~

Now we can begin eliminating terms by filling in parameters in a column, and filling in the same parameters in the remainder of the premise. We can assign the parameters to produce s(A,B) in lines 5 and 6, to produce u(A,C) in lines 2 and 6, to produce c(C,D) in lines 2, 3 and 4, to produce p(C,E) in lines 4, 7, 8, and 9, and to produce m(F,E) in lines 1, 5, 7, 8, and 9, using the theorem to eliminate terms. Further simplification can be obtained by selecting parameters to produce the same term in a column, and finally to elimination terms in the e(...) column, producing the final table:

	c(...)	e(...)	f(...)	i(...)	m(...)	p(...)	s(...)	u(...)
1)			f(E,F)~		m(F,E)
2)	c(C,D)			i(D,A)~				u(A,C)~
3)	c(C,D)~	e(K,C) e(G,K)~	f(D,G)~
4)	c(C,D)~		f(E,D)~			p(C,E)
5)			f(B,E)~	i(A,F)~	m(F,E)~		s(A,B)
6)			f(B,I)~ f(I,C)~				s(A,B)~	u(A,C)~
7)		e(J,K) e(C,K)~		i(F,J)~	m(F,E)~	p(C,E)~
8)		e(K,J) e(K,C)		i(F,J)~	m(F,E)~	p(C,E)~
9)		e(K,C)~ e(K,H)~	f(F,H)~		m(F,E)~	p(C,E)~

with the conclusion:

ρ(e(G,K)^~e(I,K)e(H,K)^~f(B,E)^~f(E,F)^~f(F,H)^~f(E,D)^~f(D,G)^~f(B,I)^~f(I,C)^~i(D,A)^~i(A,F)^~i(F,I)^~u(A,C)^~) or
e(J,K)^~ · f(B,E) · f(E,F) · f(F,H) · f(E,D) · f(D,G) · f(B,I) · f(I,C) · i(D,A) · i(A,F) · i(F,J) · u(A,C) ⇒ ¬[e(G,K) · e(H,K)]

Diagramming these relationships, we conclude that if the great-grandfather, B, had two sons E & I, and that if E had two sons F & D, who each had a son, H & G respectively, and that if I had a son C, and that if D is inferior to A who is inferior to F who is inferior to J - then if all these conditions are met, the conclusion Dodgson requested is:

The two great-grandsons (G & H) are not both enemies of their Father/Uncle's (F's) superior's (J's) friend (K).

We don't know where A fits in the family tree, only that he is an uncle of C (B's grandson); he could be either a son of B or a maternal uncle of C.  In completing the father column f(...), we assumed the existence of an individual I, a brother of E.  It is also possible that E had no brother, i.e., that E and I are the same individual.  There is not enough information from the premises to decide I's status.  None of this uncertainty affects the conclusion.
[Note: This article was corrected on 2/27/07 to fix a mistake in the table elimination and to show the new, proper conclusion.]

Quote of the Day

It is a profoundly erroneous truism, repeated by all copy books and by eminent people when they are making speeches, that we should cultivate the habit of thinking of what we are doing. The precise opposite is the case. Civilization advances by extending the number of important operations which we can perform without thinking about them.
-Alfred North Whitehead, An Introduction to Mathematics

Don't Ask Alice - Part the Seventh

This is the seventh installment on Dodgson's unsolved sorites - for the previous six go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Don't ask Alice - Part the Fourth [from Dodgson's Symbolic Logic, p. 190]
Don't ask Alice - Part the Fifth [from Dodgson's Symbolic Logic, p. 191]
Don't ask Alice - Part the Sixth [from Dodgson's Symbolic Logic, p. 192]
Without further ado, here is the seventh of Dodgson's [hitherto] unsolved sorites:

(1) Brothers, who are much admired, are apt to be self-conscious;
(2) When two men of the same height are on opposite sides in Politics, if one of them has his admirers, so also has the other;
(3) Brothers, who avoid general Society, look well when walking together;
(4) Whenever you find two men, who differ in Politics and in their views of Society, and who are not both of them ugly, you may be sure that they look well when walking together;
(5) Ugly men, who look well when walking together, are not both of them free from self-consciousness;
(6) Brothers, who differs in Politics, and are not both of them handsome, never give themselves airs;
(7) John declines to go into Society, but never gives himself airs;
(8) Brothers, who are apt to be self-conscious, though not both of them handsome, usually dislike Society;
(9) Men of the same height, who do not give them- selves airs, are free from self-consciousness;
(10) Men, who agree on questions of Art, though they differ in Politics, and who are not both of them ugly, are always admired;
(11) Men, who hold opposite views about Art and are not admited, always give themselves airs;
(12) Brothers of the same height always differ in Politics;
(13) Two handsome men, who are neither both of them admired nor both of them self-conscious, are no doubt of different heights;
(14) Brothers, who are self-conscious, and do not both of them like Society, never look well when walking together.

The universe is obviously the class of men, and there seem to be both properties (i.e., one-place functions) and relations (i.e., two-place functions) involved. Premise (7) is different from the others, being the only one to refer to a specific individual, John, and involving only properties of that individual.  Define the following relations and properties in terms of general variables X and Y: b(X,Y) – X and Y are brothers, a(X) – X is admired (has admirers), s(X) – X is self-conscious, h(X,Y) – X and Y are the same height, o(X,Y) – X and Y are opposite in politics, v(X) – X avoids general society, w(X,Y) – X and Y look well walking together, g(X) – X is handsome, r(X) – X gives himself airs, and q(X,Y) – X and Y agree on questions of art. Writing the conjunction of all the premises except (7):
ρ(([a(X)^~a(Y)^~]b(X,Y)^~[s(X)^~s(Y)^~]^~) · ρ(([a(X)a(Y)]^~[a(X)^~a(Y)^~]^~h(X,Y)^~o(X,Y)^~) · ρ((b(X,Y)^~[v(X)^~v(Y)^~]w(X,Y)) · ρ(([g(X)g(Y)]^~o(X,Y)^~[v(X)v(Y)]^~[v(X)^~v(Y)^~]^~w(X,Y)) · ρ(([g(X)g(Y)][s(X)s(Y)]w(X,Y)^~) · ρ((b(X,Y)^~[g(X)^~g(Y)^~]o(X,Y)^~[r(X)r(Y)]^~) · ρ((b(X,Y)^~[g(X)^~g(Y)^~]^~[s(X)^~s(Y)^~][v(X)^~v(Y)^~]^~) · ρ((h(X,Y)^~[r(X)r(Y)][s(X)s(Y)]^~) · ρ(([a(X)^~a(Y)^~]^~[g(X)g(Y)]^~o(X,Y)^~q(X,Y)^~) · ρ(([a(X)a(Y)]q(X,Y)[r(X)^~r(Y)^~]^~) · ρ((b(X,Y)^~h(X,Y)^~o(X,Y)) · ρ(([a(X)^~a(Y)^~]^~[g(x)^~g(Y)^~]h(X,Y)^~[s(X)^~s(Y)^~]^~) · ρ((b(X,Y)^~[s(X)^~s(Y)^~][v(X)v(Y)]^~w(X,Y))
and applying the theorem, canceling those terms which appear both complemented and uncomplemented (shown underlined above), produces the conclusion :

ρ(b(X,Y)^~h(X,Y)^~[r(X)^~r(Y)^~]^~[v(X)v(Y)]^~)

Now, from premise (7):  [v(J)^~r(J)]^~ = U, or complementing v(J)^~r(J) = U^~, whence: v(J)^~ = U^~ or v(J) = U, and r(J) = U^~. Substituting J for X in the conclusion (since the conclusion is valid for any X and therefore for John in particular) and simplifying :

ρ(b(J,Y)^~h(J,Y)^~[r(J)^~r(Y)^~]^~[v(J)v(Y)]^~) = ρ(b(J,Y)^~h(J,Y)^~[(U^~)^~r(Y)^~]^~[Uv(Y)]^~) = ρ(b(J,Y)^~h(J,Y)^~)   or    "No brothers of John are John's height."   Q.E.D.

Note that this conclusion does not necessarily entail that John has any brothers – however, if he does, then the conclusion applies to all of them.

Quote of the Day

There is no such thing as conversation. It is an illusion. There are intersecting monologues, that is all. -- Dame Rebecca West

Don't Ask Alice - Part the Sixth

This is the sixth installment on Dodgson's unsolved sorites - for the previous five go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Don't ask Alice - Part the Fourth [from Dodgson's Symbolic Logic, p. 190]
Don't ask Alice - Part the Fifth [from Dodgson's Symbolic Logic, p. 191]
Without further ado, here is the sixth of Dodgson's [hitherto] unsolved sorites:

After the six friends, named in Problem 5, had returned from their tour, three of them, Barry, Cole, and Dix, agreed with two other friends of theirs, Lang and Mill, that the five should meet, every day, at a certain table d'hôte. Remembering how much amusement they had derived from their code of rules for walking-parties, they devised the following rules to be observed whenever beef appeared on the table:---
(1) If Barry takes salt, then either Cole or Lang takes one only of the two condiments, salt and mustard: if he takes condiment, or Mill takes both.
(2) If Cole takes salt, then either Barry takes only one condiment, or Mill takes neither: if he takes mustard, then either Dix or Lang takes both.
(3) If Dix takes salt, then either Barry takes neither condiment or Cole take both: if he takes mustard, then either Lang or Mill takes neither.
(4) If Lang takes salt, then Barry or Dix takes only one condiment: if he take mustard, the either Cole or Mill takes neither.
(5) If Mill takes salt, then either Barry or Lang takes both condiments: if he takes mustard, then either Cole or Dix takes only one.
The Problem is to discover whether these rules are compatible; and, if so, what arrangements are possible.
[N.B. In this Problem, it is assumed that the phrase "if Barry takes salt" allows of two possible cases, viz. (1) "he takes salt only"; (2) "he takes both condiments". And so with all similar phrases.
It is also assumed that the phrase "either Cole or Lang takes one only of the two condiments" allows three possible cases, viz. (1) "Cole takes one only, Lang takes both or neither"; (2) "cole takes both or neither, Lang takes one only"; (3) "Cole take one only, Lang takes one only". And so with all similar phrases.
It is also assumed that every rule is to be understood as implying the words "and vice versâ." Thus the first rule would imply the addition "and, if either Cole or Lang takes only one condiment, then Barry takes salt."]

This problem is a completely different type from the previous ones, as will be seen. As in the previous problem (Problem 5), introduce the propositional characteristic functions s_J = {U,U^~} and m_J = {U,U^~} representing the condition (true or false, respectively) that the individual "J" uses salt or mustard, respectively. The universe of individuals is {B, C, D, L, M} representing Barry, Cole, Dix, Lang, and Mill, so there are 10 unknowns in all. From Dodgson's comments following the premises, each premise is not simply an implication but a material equivalence, since implications are assumed in both directions. So the premises can be transcribed as:
1a) s_b = (s_c^ m_c)(s_l^ m_l)
1b) m_b = (s_d^~ · m_d^~)(s_m · m_m)
2a) s_c = (s_b^ m_b)(s_m^~ · m_m^~)
2b) m_c = (s_d · m_d)(s_l · m_l)
3a) s_d = (s_b^~ · m_b^~)(s_c · m_c)
3b) m_d = (s_l^~ · m_l^~)(s_m^~ · m_m^~)
4a) s_l = (s_b^ m_b)(s_d^ m_d)
4b) m_l = (s_c^~ · m_c^~)(s_m^~ · m_m^~)
5a) s_m = (s_b · m_b)(s_l · m_l)
5b) m_m = (s_c^ m_c)(s_d^ m_d)
where, following the boolean conventions employed in this series, concatenation represents disjunction (or), "^~" represents complementation (negation), "·" represents conjunction (and), and "^" represents the exclusive-or operation.

Now, there are no terms appearing complemented and uncomplemented, and therefore the general method used in the previous problems is not applicable. The arithmetic approach of traditional propositional logic would enumerate all combinations of the 10 variables, resulting in 2¹⁰ = 1,024 combinations, and then proceed to determine which, if any, satisfied the 10 relations (1a)-(5b). This is so unwieldy that no one would even attempt it. The obvious algebraic approach is substitution to successively eliminate unknowns. Substituting (1a) and (1b) into (2a):
(2a.1) s_c = (s_b^ m_b)(s_m^~ · m_m^~) = {[(s_c^ m_c)(s_l^ m_l)]^~(s_d^~· m_d^~)(s_m · m_m)}^~ {(s_c^ m_c)(s_l^ m_l)[(s_d^~· m_d^~)(s_m · m_m)]^~}^~ (s_m^~· m_m^~)
This is an equation in which the variable s_c appears on both sides of the equality. Simplifying and collecting terms in s_c and s_c^~ on the right side:
(2a.2) s_c = (s_m^~· m_m^~) [(s_d^~· m_d^~)(s_l^ m_l)^~(s_m· m_m)]^~ [(s_d^ m_d)^~(s_l· m_l)(s_m· m_m)s_c^~]^~ [{[(s_d· m_d)^~(s_m· m_m)]^~ [(s_d^~· m_d^~)(s_l^ m_l)]^~ [(s_d^ m_d)^~(s_l^~· m_l^~)^~]^~ [(s_l^ m_l)(s_m· m_m)^~]^~ [(s_d^~· m_d^~)(s_l· m_l)^~(s_m· m_m)]^~ [(s_d· m_d)(s_l^~· m_l^~)^~(s_m· m_m)^~]^~}^~s_c]^~
We seem to have reached an impasse – this type of equation is outside the realm of traditional logic. Fortunately, there is a simple method available in this case. Since the variables are assumed propositional, we can assume the two possible values for s_c and investigate the consequences of each assumption.

First, assume s_c = U^~. Substituting this value in (2a.2):
(I.2a.3)   U^~ = (s_m^~· m_m^~) [(s_d^~· m_d^~)(s_l^ m_l)^~(s_m· m_m)]^~ [(s_d· m_d)^~(s_m· m_m)]^~ [(s_d^~· m_d^~)(s_l^ m_l)]^~ [(s_d^ m_d)^~(s_l^~· m_l^~)^~]^~ [(s_l^ m_l)(s_m· m_m)^~]^~ [(s_d^~· m_d^~)(s_l· m_l)^~(s_m· m_m)]^~ [(s_d· m_d)(s_l^~· m_l^~)^~(s_m· m_m)^~]^~
from which we can conclude, using using an inference theorem and complementing if necessary:
(I.6a)   (s_m^~· m_m^~) = U^~
(I.6b)   (s_d^~· m_d^~)(s_l^ m_l)^~(s_m· m_m) = U
(I.6c)   (s_d· m_d)^~(s_m· m_m) = U
(I.6d)   (s_d^~· m_d^~)(s_l^ m_l) = U
(I.6e)   (s_d^ m_d)^~(s_l^~· m_l^~)^~ = U
(I.6f)   (s_l^ m_l)(s_m· m_m)^~ = U
(I.6g)   (s_d^~· m_d^~)(s_l· m_l)^~(s_m· m_m) = U
(I.6h)   (s_d· m_d)(s_l^~· m_l^~)^~(s_m· m_m)^~ = U
Applying the theorem to (I.6b) and (I.6d), (s_d^~· m_d^~)(s_m· m_m) = U, whence m_b = U. We can now start substituting and reducing the other relations:
(I.3a.1)   s_d = (s_b^~· m_b^~)(s_c· m_c) = (s_b(U^~)^~)^~((U^~)^~m_c^~)^~ = U^~
(I.2b.1)   m_c = (s_d· m_d)(s_l· m_l) = (U^~· m_d)(s_l· m_l) = (s_l· m_l)
(I.1a.1)   s_b = (s_c^ m_c)(s_l^ m_l) = (U^~^ m_c)(s_l^ m_l) = m_c(s_l^ m_l)= (s_l^ m_l)(s_l· m_l) = (s_l^~m_l)^~(s_lm_l^~)^~(s_l^~m_l^~)^~ = s_lm_l
(I.3b.1)   m_d = (s_l^~· m_l^~)(s_m^~· m_m^~) = (s_l^~· m_l^~)U^~ = (s_l^~· m_l^~)
(I.4a.1)   s_l = (s_b^ m_b)(s_d^ m_d) = (s_b^ U)(U^~^ m_d) = s_b^~m_d = (s_lm_l)^~
The only consistent combination satisfying this equation is s_l = U^~ and m_l = U. Finishing the substitutions:
(I.2b.2)   m_c = (s_l· m_l) = ((U^~)^~U^~)^~ = U^~
(I.1a.2)   s_b = s_lm_l = U^~U = U
(I.3b.2)   m_d = (s_l^~· m_l^~) = (((U^~)^~)^~(U^~)^~)^~ = U^~
(I.5a.1)   s_m = (s_b· m_b)(s_l· m_l) = (U^~U^~)^~((U^~)^~U^~)^~ = UU^~ = U
(I.5b.1)   m_m = (s_c^ m_c)(s_d^ m_d) = ((U^~)^~U^~)^~(U^~(U^~)^~)^~((U^~)^~U^~)^~(U^~(U^~)^~)^~ = U^~
Substitution of these values show that (1a)-(5b) are all satisified. Hence, all variables are completely determined and there is only one possible set of values for this assumption.

If we instead assume s_c = U, then we eventually encounter a contradiction, so the detailed calculations will not be presented. We can now definitively and conclusively answer Dodgson's questions, towit: The rules are compatible, and there is only one possible arrangement:

Barry and Mill take salt, and Barry and Lang take mustard.

This is not unreasonable; for a system of 10 independent, consistent equations in 10 unknowns, only one solution would be expected.

The other question, for which there is no answer, is: what was Dodgson's solution for this problem? The solution given here has used algebraic methods which were unavailable to Dodgson and are unususal even in modern symbolic logic. I have been unable to obtain any solution using the methods in Dodgson's book, and therefore am unable to envision Dodgson's method. The fact that he used the plural 'arrangements' might indicate he thought there was more than one solution, but this is mere speculation.

Quote of the Day

For in much wisdom is much grief; and he that increaseth knowledge increaseth sorrow. - Ecclesiastes 1:18 -

Don't Ask Alice - Part the Fifth

This is the fifth installment on Dodgson's unsolved sorites - for the previous four go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Don't ask Alice - Part the Fourth [from Dodgson's Symbolic Logic, p. 190]
Without further ado, here is the fifth of Dodgson's [hitherto] unsolved sorites:

Six friends, and their six wives, are staying in the same hotel; and they all walk out daily, in parties of various size and composition. To ensure variety in these daily walks, they have agree to observe the following Rules:---
(1) If Acres is with (i.e. is in the same party with) his wife, and Barry with his, and Eden with Mrs. Hall, Cole must be with Mrs. Dix;
(2) If Acres is with his wife, and Hall with his, and Barry with Mrs. Cole, Dix must not be with Mrs. Eden;
(3) If Cole and Dix and their wives are all in the same party, and Acres not with Mrs. Barry, Eden must not be with Mrs. Hall;
(4) If Acres is with his wife, and Dix with his, and Barry not with Mrs. Cole, Eden must be with Mrs. Hall;
(5) If Eden is with his wife, and Hall with his, and Cole with Mrs. Dix, Acres must not be with Mrs. Barry;
(6) If Barry and Cole and their wives are all in the same party, and Eden not with Mrs. Hall, Dix must be with Mrs. Eden.
The Problem is to prove that there must be, every day, at least one married couple who are not in the same party.

This problem is simplest by introducing the propositional functions g_J(I) = {U,U^~} representing the condition (true or false) that the individual I is in the same party as Mrs. J (this type of function was unknown to Dodgson). The universe of individuals is {A, B, C, D, E, H, MA, MB, MC, MD, ME, MH} representing Acres, Barry, Cole, Dix, Eden, Hall, and Mrs. Acres, etc. (the Mrs. don't enter into the problem except through the functions). Transcribing each of the premises and writing them as a conjunction:
ρ(g_A(A)^~g_B(B)^~g_H(E)^~g_D(C)) · ρ(g_A(A)^~g_H(H)^~g_C(B)^~g_E(D)^~) · ρ(g_C(C)^~g_D(D)^~g_B(A)g_H(E)^~) · ρ(g_A(A)^~g_D(D)^~g_C(B)g_H(E)) · ρ(g_E(E)^~g_H(H)^~g_D(C)^~g_B(A)^~) · ρ(g_B(B)^~g_C(C)^~g_H(E)g_E(D))
Applying the theorem, canceling those terms which appear both complemented and uncomplemented (shown underlined above), produces the conclusion:

ρ(g_A(A)^~g_B(B)^~g_C(C)^~g_D(D)^~g_E(E)^~g_H(H)^~),    or
¬(g_A(A) · g_B(B) · g_C(C) · g_D(D) · g_E(E) · g_H(H))

Hence, that all of the men are in groups with their wives must be false, or at least one married couple is not in the same party. Q.E.D.

Quote of the Day

I tell them that if they will occupy themselves with the study of mathematics, they will find in it the best remedy against the lusts of the flesh. -- Thomas Mann, The Magic Mountain --

Don't Ask Alice - Part the Fourth

This is the fourth installment on Dodgson's unsolved sorites - for the first three go here:
Don't ask Alice, I don't think she'll know [from Dodgson's Symbolic Logic, p. 186]
Don't ask Alice - Part the Second [from Dodgson's Symbolic Logic, p. 187]
Don't ask Alice - Part the Third [from Dodgson's Symbolic Logic, p. 188]
Without further ado, here is the fourth of Dodgson's [hitherto] unsolved sorites:

(1) Any one, fit to be an M.P., who is not always speaking, is a public benefactor;
(2) Clear-headed people, who express themselves well have had a good education;
(3) A woman, who deserves praise, is one who can keep a secret;
(4) People, who benefit the public, but do not use their influence for good purpose, are not fit to go into Parliament;
(5) People, who are worth their weight in gold and who deserve praise, are always unassuming;
(6) Public benefactors, who use their influence for good objects, deserve praise;
(7) People, who are unpopular and not worth their weight in gold, never can keep a secret;
(8) People, who can talk for ever and are fit to be Members of Parliament, deserve praise;
(9) Any one, who can keep a secret and who is unassuming, is a never-to-be-forgotten public benefactor;
(10) A woman, who benefits the public, is always popular;
(11) People, who are worth their weight in gold, who never leave off talking, and whom it is impossible to forget, are just the people whose photographs are in all the shop-windows;
(12) An ill-educated woman, who is not clear-headed, is not fit to go into Parliament;
(13) Any one, who can keep a secret and is not for ever talking, is sure to be unpopular;
(14) A clear-headed person, who has influence and uses it for good objects, is a public benefactor;
(15) A public benefactor, who is unassuming, is not the sort of person whose photograph is in every shop-window;
(16) People, who can keep a secret and who use their influence for good purposes, are worth their weight in gold;
(17) A person, who has no power of expression and who cannot influence others, is certainly not a woman;
(18) People, who are popular and worthy of praise, either are public benefactors or else are unassuming.
Univ. "persons"; a = able to keep a secret; b = clear-headed; c = constantly talking; d = deserving praise; e = exhibited in shop-windows; h = expressing oneself well; k = fit to be an M.P.; l = influential; m = never-to-be-forgotten; n = popular; r = public benefactors; s = unassuming; t = using one's influence for good objects; v = well-educated; w = women; z = worth one's weight in gold.

The solution proceeds exactly as the first three problems. It will be seen in what follows that premise 12 is redundant; premise 9 can be split into two simpler premises and writing the premises except 12 as a conjunction:
ρ(CK^~R) · ρ(B^~H^~V) · ρ(AD^~W^~) · ρ(K^~R^~T) · ρ(D^~SZ^~) · ρ(DR^~T^~) · ρ(A^~NZ) · ρ(C^~DK^~) · ρ(A^~MS^~) · ρ(A^~RS^~) · ρ(NR^~W^~) · ρ(C^~EM^~Z^~) · ρ(A^~CN^~) · ρ(B^~L^~RT^~) · ρ(E^~R^~S^~) · ρ(A^~T^~Z) · ρ(HLW^~) · ρ(D^~N^~RS)
Applying the theorem, canceling those terms which appear both complemented and uncomplemented (shown underlined above), produces the conclusion:

ρ(B^~K^~VW^~),    or 'Clear-headed women fit to be M.P.s are well-educated.' Q.E.D.

It is clear that if premise (12), ρ(K^~VW^~), is included, it becomes the minimal conclusion, with the conclusion above derivable from it, rendering all the other premises irrelevant. Dodgson apparently erred by including this premise in the group - he was guilty of "begging-the-question".

Quote of the Day

The name of the song is called "HADDOCKS' EYES."'
`Oh, that's the name of the song, is it?' Alice said, trying to feel interested.
`No, you don't understand,' the Knight said, looking a little vexed. `That's what the name is CALLED. The name really IS "THE AGED AGED MAN."'
`Then I ought to have said "That's what the SONG is called"?' Alice corrected herself.
`No, you oughtn't: that's quite another thing! The SONG is called "WAYS AND MEANS": but that's only what it's CALLED, you know!'
`Well, what IS the song, then?' said Alice, who was by this time completely bewildered.
`I was coming to that,' the Knight said. `The song really IS "A-SITTING ON A GATE": and the tune's my own invention.'
- Lewis Carroll, Alice Through the Looking Glass

Don't Ask Alice - Part the Third

This is the third installment on Dodgson's unsolved sorites - for the first two go here:
Don't ask Alice, I don't think she'll know
Don't ask Alice - Part the Second
Without further ado, here is the third of Dodgson's [hitherto] unsolved sorites:

(1) When the day is fine, I tell Froggy "You're quite the dandy, old chap!";
(2) Whenever I let Froggy forget that £10 he owes me, and he begins to strut about like a peacock, his mother declares "He shall not go out a-wooing!";
(3) Now that Froggy's hair is out of curl, he has put away his gorgeous waistcoat;
(4) Whenever I go out on the roof to enjoy a quiet cigar, I'm sure to discover that my purse is empty;
(5) When my tailor calls with his little bill, and I remind Froggy of that £10 he owes me, he does not grin like a hyæna;
(6) When it is very hot, the thermometer is high;
(7) When the day is fine, and I'm not in the humour for a cigar, and Froggy is grinning like a hyæna, I never venture to hint that he's quite the dandy;
(8) When my tailor calls with his little bill and finds me with an empty purse, I remind Froggy of that £10 he owes me;
(9) My railway-shares are going up like anythin!
(10) When my purse is empty, and when, noticing that Froggy has got his gorgeous waistcoat on, I venture to remind him of that £10 he ower me, things are apt to get rather warm;
(11) Now that it looks like rain, and Froggy is grinning like a hyæna, I can do without my cigar;
(12) When the thermometer is high, you need not trouble yourself to take an umbrella;
(13) When Froggy has his gorgeous waistcoat on, but is not strutting about like a peacock, I betake myself to a quiet cigar;
(14) When I tell Froggy that he's quite the dandy, he grins like a hyæna;
(15) When my purse is tolerably full, and Froggy's hair is one mass of curls, and when he is not strutting about like a peacock, I go out on the roof;
(16) When my railway-shares are going up, and when it is chilly and looks like rain, I have a quiet cigar;
(17) When Froggy's mother lets him go a-wooing, he seems nearly mad with joy, and puts on a waistcoat that is gorgeous beyond words;
(18) When it is going to rain, and I am having a quiet cigar, and Froggy is not intending to go a-wooing, you had better take an umbrella;
(19) When my railway-shares are going up, and Froggy seems nearly mad with joy, that is the time my tailor always chooses for calling with his little bill;
(20) When the day is cool and the thermometer low, and I say nothing to Froggy about his being quite the dandy, and there's not the ghost of a grin on his face, I haven't the heart for my cigar!

The solution proceeds exactly as the first two problems. Define the following propositional statements: R – 'I go out on the roof.', Q – 'I enjoy a quiet cigar.', P – 'My purse is empty.', T – 'My tailor calls with his little bill.', O – 'I remind Froggy of the £10 he owes me.', D – 'I tell Froggy "You're quite the dandy, old chap."', S – 'Froggy struts about like a peacock.', A – 'Froggy goes out a-wooing.', C – 'Froggy's hair is curly.', W – 'Froggy wears his gorgeous waistcoat.', G – 'Froggy grins like a hyena.', J – 'Froggy seems nearly mad with joy.', C – 'The day is cool.', H – 'The day is hot.', M – 'The thermometer is high.', L – 'It looks like rain.', and B – 'You had better take an umbrella.'. Interpret the statement 'The day is fine.' as meaning it is neither cool nor hot, or (CH)^~. Since premise (9) tells us that railroad-shares are going up, it can be omitted and this condition in statements (16) and (19) can be ignored. Finally, premise 17 can be expanded into two simpler statements. Using these propositional statements, write the premises as the conjunction:
ρ(CDH) · ρ(A^~OS^~) · ρ(CW^~) · ρ(PQ^~R^~) · ρ(G^~O^~T^~) · ρ(H^~M) · ρ(CG^~HO^~Q) · ρ(OP^~T^~) · ρ(HO^~P^~W^~) · ρ(G^~L^~Q^~) · ρ(B^~M^~) · ρ(GQW^~) · ρ(D^~G) · ρ(C^~PRS) · ρ(C^~L^~Q) · ρ(A^~J) · ρ(A^~W) · ρ(ABL^~Q^~) · ρ(J^~T) · ρ(C^~DGMQ^~)
Applying the theorem, canceling those terms which appear both complemented and uncomplemented (shown underlined above), produces the conclusion:

ρ(L^~),    or 'It doesn't look like it will rain.' Q.E.D.

Again, what seems to be a very complex, seemingly impossible problem turns out to have a simple solution which is easily found.