Who becomes king?

Question

5 earls argue which becomes king and which becomes treasurer.

A will be happy only if D or E is treasurer.

B will be happy only if C is treasurer.

C will be happy only if D is either king or treasurer.

D will be happy only if A is king or treasurer.

E will be happy only if A is not king.

It is not possible for all five to be happy,so in the end they appoint king and treasurer so that the other 3 will be happy.

Forming a table,I find that D will be treasurer since A,C and E will be happy.So that eliminates D as king.Then,how can I proceed further since only C,D and E states who they want to be king.

Edit:I forgot to type in part of the question.

Answer 1

We start with

$$\begin{array} \\ \tag{1} a &:& t(d) \lor t(e) \\ b &:& t(c) \\ c &:& k(d) \lor t(d) \\ d &:& k(a) \lor t(a) \\ e &:& \lnot k(a) \end{array}$$

So we have five statements (a,b,c,d,e). All statements except the statement of the king and the treasurer must be true. We have some additional conditions $(2)$ that must be true:

• there is exactly one treasurer
• there exactly one king
• the king cannot be the treasurer

To simplify the processing of the statements $81)$ we augment each statement $x$ ($x \in \{a,b,c,d,e\}$) by $t(x) \lor k(x)$ .

Now we can search for a solution such that $(2)$ and all filve statements in $(3)$ are true.

$$\begin{array} \\ \tag{3} a &:& t(d) \lor t(e) \lor k(a) \lor t(a) \\ b &:& t(c) \lor k(b) \lor t(b) \\ c &:& k(d) \lor t(d) \lor k(c) \lor t(c) \\ d &:& k(a) \lor t(a) \lor k(d) \lor t(d) \\ e &:& \lnot k(a) \lor k(e) \lor t(e) \end{array}$$

We reorder the terms $$\begin{array} \\ a &:& k(a) \lor t(a) \lor t(d) \lor t(e) \\ \tag{4} b &:& k(b) \lor t(b) \lor t(c) \\ c &:& k(d) \lor k(c) \lor t(c) \lor t(d) \\ d &:& k(a) \lor k(d) \lor t(a) \lor t(d) \\ e &:& k(b) \lor k(c) \lor k(d) \lor k(e) \lor t(e) \end{array}$$

and see that it statement $a$ and statement $b$ are both true then $k(a) \lor k(b)$ must be true ans so we have

$$\lnot k(c), \lnot k(d), \lnot k(e) \tag{5}$$

If we apply $(5)$ on $(4)$ we get:

$$\begin{array} \\ \tag{6} a &:& k(a) \lor t(a) \lor t(d) \lor t(e) \\ b &:& k(b) \lor t(b) \lor t(c) \\ c &:& t(c) \lor t(d) \\ d &:& k(a) \lor t(a) \lor t(d) \\ e &:& k(b) \lor t(e) \end{array}$$

and from statement $c$ in $(6)$ we conclude

$$\lnot t(a), \lnot t(b), \lnot t(e) \tag{7}$$

$(7)$ applied to $(6)$ gives

$$\begin{array} \\ \tag{8} a &:& k(a) \lor t(d) \\ b &:& k(b) \lor t(c) \\ c &:& t(c) \lor t(d) \\ d &:& k(a) \lor t(d) \\ e &:& k(b) \end{array}$$

From statement $e$ and $a$ and $(2)$ follows $t(d)$. $k(b)$ and $t(d)$ satisfy all statements of $(8)$ (and therefore $(3)$)and $(2)$.