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Location: Chapter 22, additional questions after Problem 17 and before "Kestrels and infinity"



For several birds, it is suggested to try and prove their nonegocentricity. In the following, the birds listed in the following are considered:

\forall \beta \in \{ D, E, F, G, H, L, J, O, Q_1, Q_2, Q_3, Q_4, W', C^\star, C^{\star\star}, W^\star, W^{\star\star}, U \}~~\Rightarrow~~\beta\beta\neq\beta\;.


It is also suggested to prove that any two distinct birds taken from \{B,C,W,S,R,T\} are not even similar (i.e. they can always be distinguished by their action on some birds).

Considering the birds in the set are all proper birds of order \leq 3, proving this amounts to proving that

\forall \beta_1, \beta_2 \in \{B,C,W,S,R,T\}, \beta_1\neq\beta_2 \Rightarrow \beta_1 xyz \neq \beta_2 xyz\;.

(In an extensional, or sparse, forest, this amounts to the statement that the six birds in the above set are all different).



For each bird \beta, the nonegocentricity is proven by first assuming that \beta\beta=\beta, and then by deriving from the assumption some contradiction involving Kestrels and Identity birds such as K=KK, K=I. To get to the contradiction, the assumption is applied to two, three or more variables and then values \in \{K,I\} are assigned to some of them.


It is Dxyzw=xy(zw). Starting from DD=D, it is

Dxyzw=DDxyzw \to xy(zw)v=Dx(yz)wv \to xy(zw)v=x(yz)(wv)\;.

With x=v=K, y=z=I:

KI(Iw)K=K(II)(wK) \to IK=II \to K=I\;,

a contradiction (Problem 2).


It is Exyzwv=xy(zwv). If EE=E,

EExyzwv=Exyzwv \to Ex(yzw)vqr=xy(zwv)qr \to x(yzw)(vqr)=xy(zwv)qr\;.

With x=w=K and y=z=q=r=I:

K(IIK)(vII)=KI(IKv)II \to IIK=III \to K=I\;,

a contradiction (Problem 2).


It is Fxyz=zyx. From FF=F,

FFxyz=Fxyz \to yxFzwv=zyxwv \to yxvwz=zyxwv\;,

then setting x=y=z=I, w=v=K:


another contradiction (Problem 6).


It is Gxyzw=xw(yz). If GG=G,

GGxyzw=Gxyzw \to Gz(xy)wq=xw(yz)q \to zq(xyw)=xw(yz)q \;,

then when x=z=w=K and q=I:

KI(KyK)=KK(yK)I \to I=KI\;,

contradicting Problem 1.


It is Hxyz=xyzy. Now, HH=H implies

HHxyz=Hxyz \to Hxyxz=xyzy \to xyxyz=xyzy\;;

setting x=z=K, y=I one finds

KIKIK=KIKI \to IIK=II \to K=I\;,

which is against Problem 2.


Since Lxy=x(yy), if one had LL=L it would follow that

LLxy=Lxy \to L(xx)y=x(yy) \to xx(yy)=x(yy)\;,

and then immediately, with x=K and y=I,

KK(II)=K(II) \to K=KI\;,

at odds with Problem 6.


It is Jxyzw=xy(xwz). From JJ=J:

\begin{align} JJxyzw= & Jxyzw \to Jx(Jzy)wq=xy(xwz)q \to x(Jzy)(xqw)=xy(xwz)q \to \\ \stackrel{x=K}{\longrightarrow} & K(Jzy)(kqw)=Ky(Kwz)q \to Jzy\alpha\beta=yz\alpha\beta \to zy(z\beta\alpha) = yq\alpha\beta\;; \end{align}

finally, setting y=z=K, q=\alpha=\beta=I, one gets

KK(KII)=KIII \to K=I\,,

i.e. a contradiction of Problem 2.


It is Oxy=y(xy). The assumption OO=O implies:

\begin{align} OOxy=Oxy \to & x(Ox)y=y(xy) \stackrel{x=y=K}{\longrightarrow} K(OK)K=K(KK) \to OKz=K(KK)z \to \\ \stackrel{z=I}{\longrightarrow} & I(KI)=KK \to K=I\;, \end{align}

(where the left cancellation law for Kestrels has been used) which goes against Problem 2.

Quixotic bird

Since Q_1xyz=x(zy), from Q_1Q_1=Q_1 one gets:

Q_1xyz=Q_1Q_1xyz \to x(zy)w=Q_1(yx)zw \to x(zy)w=yx(wz)\;,

i.e., setting x=y=z=K,

K(KK)w=KK(wK) \to KK=K\;,

thereby contradicting Problem 6.

Quizzical bird

It is Q_2xyz=y(zx), hence from Q_2Q_2=Q_2 one finds:

Q_2Q_2xyz=Q_2xyz \to x(yQ_2)z=y(zx)\;,

which choosing x=K and y=z=I becomes

K(IQ_2)I=I(IK) \to Q_2=K\;.

Applying the last identity to a,b,c and then specialising to a=c=I, b=K one finds:

Q_2abc=Kabc \to b(ca)=ac \to K(II)=II \to KI=I\;,

a contradiction of Problem 1.

Quirky bird

Since Q_3xyz=z(xy), in case of egocentricity one would have

Q_3Q_3xyz=Q_3xyz \to y(Q_3x)z=z(xy)\;,

i.e., when x=I and y=z=K,

K(Q_3I)K=K(IK) \to Q_3Iab=KKab \to ba=Kb\;;

finally, the choice a=b=I leads to I=KI, again contradicting Problem 1.

Quacky bird

From the definition Q_4xyz=z(yx), Q_4Q_4=Q_4 implies

\begin{align} Q_4Q_4xyz=Q_4xyz \to & y(xQ_4)z=z(yx) \stackrel{y=K}{\longrightarrow} K(xQ_4)z=z(Kx) \stackrel{x=I}{\longrightarrow} Q_4=z(KI) \\ \stackrel{z=K}{\longrightarrow} & Q_4abc=K(KI)abc \to c(ba)=KIbc \;, \end{align}

which choosing a=b=I, c=K yields the statement K=KI, incompatible with Problem 6.

Converse warbler

From W'xyz=yxx, the assumption W'W'=W' leads to

\begin{align} W'W'xyz=W'xyz \to & xW'W'y=yxx \stackrel{x=I, y=K}{\longrightarrow} KW'W' = I \\ \to & W'ab=Iab \to baa=ab \stackrel{a=I, b=K}{\longrightarrow} I=K\;, \end{align}

a contradiction of Problem 2.

Cardinal once removed

Since C^\star xyzw=xywz, if the bird were egocentric it would be:

C^\star C^\star xyzw = C^\star xyzw \to C^\star xzyw=xywz \to xzwy=xywz\;,

then, choosing x=y=w=I and z=K one gets IKII=IIIK, i.e. I=K, incompatible wiith Problem 2.

Cardinal twice removed

It is C^{\star\star} xyzwv=xyzvw; assuming egocentricity implies

C^{\star\star} C^{\star\star} xyzwv=C^{\star\star} xyzwv \to C^{\star\star} xywzv=xyzvw \to xywvz=xyzvw\;,

which choosing x=y=z=v=I and w=K yields


contradicting Problem 2.

Warbler once removed

It is W^\star xyz=xyzz; now, if W^\star W^\star =W^\star one would have

W^\star W^\star xyz=W^\star xyz \to W^\star xyyz=xyzz \to xyyyz=xyzz\;;

by setting x=z=I and y=K this is seen to lead to


a contradiction of Problem 1.

Warbler twice removed

Since W^{\star\star} xyzw=xyzww, egocentricity would imply:

W^{\star\star} W^{\star\star} xyzw=W^{\star\star} xyzw \to W^{\star\star} xyzzw=xyzww \to xyzzzw=xyzww\;,

which choosing x=y=z=I and w=K yields


a statement at odds with Problem 6.

Turing bird

Since Uxy=y(xxy), assuming UU=U would lead to:

UUxy=Uxy \to x(UUx)y=y(xxy) \stackrel{x=K}{\longrightarrow} K(UUK)y=y(KKy) \to UUK=yK\;;

by taking y=I and y=K one gets respectively UUK=K and UUK=KK, hence K=KK in contradiction with Problem 6.

Note: similar proofs would hold for other birds, but see the related open problems for less obvious cases.


For each pair (\beta_1,\beta_2) of birds, the fact that they are distinct will be proven according to the following procedure:

  • assumption of identity \beta_1=\beta_2;
  • application of their identity to three generic birds: \beta_1 xyz=\beta_2 xyz;
  • assignment of particular values \in \{K,I\} to the birds x,y,z;
  • a paradox, i.e. a contradiction of some of the results found throughout the Chapter, will be found, which invalidates the original identity assumption.

The action of the birds considered on the birds xyz is the following:

\begin{align} Bxyz = & \; x(yz) \;, \\ Cxyz = & \; xzy \;, \\ Wxyz = & \; xyyz\;, \\ Sxyz = & \; xz(yz)\;, \\ Rxyz = & \; yzx\;, \\ Txyz = & \; yxz\;. \\ \end{align}

Cardinal & Bluebird

If Cxyz=Bxyz, then xzy=x(yz); hence, with x=y=I and z=K, one has:

IKI=I(IK) \to KI=K\;,

contradicting Problem 6.

Note: other choices are also possible for most of these proofs. In this case, for instance, the assignment x=z=K and y=I would lead to KKI=K(IK) \to K=KK, against a contradiction of Problem 6.

Warbler and Bluebird

Wxyz=Bxyz \to xyyz=x(yz)\;.

Taking x=I, y=z=K:

IKKK=I(KK) \to K=KK\;,

against Problem 6.

Starling and Bluebird

Sxyz=Bxyz \stackrel{x=z=K, y=I}{\longrightarrow} KK(IK)=K(IK) \to K=KK\;\nLeftrightarrow\mbox{Prob. 6}

Robin and Bluebird

Rxyz=Bxyz \stackrel{x=I, y=z=K}{\longrightarrow} KKI=I(KK) \to K=KK\;\nLeftrightarrow\mbox{Prob. 6}

Thrush and Bluebird

Txyz=Bxyz \stackrel{x=y=z=K}{\longrightarrow} KKK=K(KK) \to K_1=K_3\;\nLeftrightarrow\mbox{Prob. 19}

Warbler and Cardinal

Wxyz=Cxyz \stackrel{x=y=z=K}{\longrightarrow} KKKK=KKK \to K=KK\;\nLeftrightarrow\mbox{Prob. 6}

Starling and Cardinal

Sxyz=Cxyz \stackrel{x=z=I, y=K}{\longrightarrow} II(KI)=IIK \to KI=K\;\nLeftrightarrow\mbox{Prob. 6}

Robin and Cardinal

Rxyz=Cxyz \stackrel{x=z=K, y=K}{\longrightarrow} IKK=KKI \to KK=K\;\nLeftrightarrow\mbox{Prob. 6}

Thrush and Cardinal

Txyz=Cxyz \stackrel{x=y=I, z=K}{\longrightarrow} IIK=IKI \to K=KI\;\nLeftrightarrow\mbox{Prob. 6}

Starling and Warbler

Sxyz=Wxyz \stackrel{x=y=z=K}{\longrightarrow} KK(KK)=KKKK \to K=KK\;\nLeftrightarrow\mbox{Prob. 6}

Robin and Warbler

Rxyz=Wxyz \stackrel{x=y=I, z=K}{\longrightarrow} IKI=IIIK \to KI=K\;\nLeftrightarrow\mbox{Prob. 6}

Thrush and Warbler

Txyz=Wxyz \stackrel{x=z=K, y=I}{\longrightarrow} IKK=KIIK \to KK=K\;\nLeftrightarrow\mbox{Prob. 6}

Robin and Starling

Rxyz=Sxyz \stackrel{x=K, y=z=I}{\longrightarrow} IIK=KI(II) \to K=I\;\nLeftrightarrow\mbox{Prob. 2}

Thrush and Starling

Txyz=Sxyz \stackrel{x=K, y=z=I}{\longrightarrow} IKI=KI(II) \to KI=I\;\nLeftrightarrow\mbox{Prob. 6}

Thrush and Robin

Txyz=Rxyz \stackrel{x=y=I, z=K}{\longrightarrow} IIK=IKI \to K=KI\;\nLeftrightarrow\mbox{Prob. 6}