Definition 3.4.1. Ani-filterof a multi-Boolean algebraA= (A,∨i,¬ij,⊥i)i,j∈A
is a nonempty subsetF ofA satisfying
(F1)⊤i∈F,
(F2) ifa, b∈F then a∨jb∈F for allj 6=i,
(F3) ifa∈F andaib thenb∈F.
An i-filter is called proper if ⊥i ∈/ F. An i-prime filter is a proper i-filter
such that
(F4) for every pair of elementsa, b∈A,a∨ib∈F implies a∈F orb∈F.
The collection ofi-filters ofAis denoted withFi(A). WithFip(A) we denote
the collection of i-prime filters of A. Next, we will define the concept of an i-(prime) ideal.
Definition 3.4.2. Ani-idealof a multi-Boolean algebraA= (A,∨i,¬ij,⊥i)i,j∈A
is a nonempty subsetI ofA satisfying
(I1)⊥i∈I,
(I2) ifa, b∈I then a∨ib∈I,
(I3) ifa∈I andbiathenb∈I.
An i-ideal is called proper if ⊤i ∈/ I. An i-prime ideal is a proper i-ideal
such that
(I4) For every pair of elementsa, b∈A,a∨jb∈I for somej 6=i∈Aimplies
a∈I orb∈I
The collection ofi-ideals ofAis denoted withIi(A). WithIip(A) we denote
the collection ofi-prime ideals of A. It is important to observe that given an arbitraryi-filterF ofA= (A,∨i,¬ij,⊥i)i,j∈A and two elementsa, b∈A
a=ibimpliesa∈F iffb∈F. (3.1)
Remember that a=i b is shorthand notation forai b andb i a. Thus
wheneveraorbare inF it follows immediately from (F3) that the other element is inF as well. The same also holds for i-ideals. Note that, contrary to the standard ‘two-player’ filters and ideals,i-ideal andi-filters are not order-duals. That is, we cannot exchange⊤i with⊥i and∨iwith some∨j in the conditions
ofi-filters to obtain the conditions ofi-ideals. Because of this fact, we will not be able to restrict ourselves to the use ofi-filters or i-ideals only, we will need both concepts to prove our desired representation theorem.
Just like in case of classical ideals and filters, the complement of an (proper) i-filter is not necessarily an (proper) i-ideal. However, the complement of an i-primeideal is ani-primefilter. We will prove this fact in the next proposition.
Proposition 3.4.3. For any multi-Boolean algebra A = (A,∨i,¬ij,⊥i)i,j∈A,
I⊂A is ani-prime ideal if and only ifA\I is ani-prime filter.
Proof. For the direction from left to right, assume that I ⊂ A is and i-prime ideal. We have to show thatF =A\Idoes not contain ⊥i and satisfies (F1)-
(F4). (F1) follows immediately from the assumption thatI is proper. Let us have a look at (F2). Suppose thata, b∈F. We need to show thata∨jb ∈F
for allj 6=i. Suppose for contradiction that for some j 6=i, a∨jb /∈F. This
implies thata∨jb∈I. But sinceI is assumed to be a proper i-prime filter it
follows from (I4) that a ∈I or b ∈I, which contradicts our assumption that bothaandbare inF. In a similar fashion, (F3) follows from (I3) and (F4) can be proved using (I2). By our assumption thatI is ani-ideal and thus contains
⊥i by definition, it follows thatF is proper.
The proof for the other direction is similar. (I1) follows from the fact thatF is proper, (I2) from (F4), (I3) from (F3) and (I4) from (F2). Also,I is proper by (F1).
Lemma 3.4.4. Let A= (A,∨i,¬ij,⊥i)i,j∈A be a multi-Boolean algebra. Let F
be ani-filter ofAanda an element ofF. Then for all b∈A a∨ib∈F. Also,
froma∨jb∈F it follows that both a andbare in F.
Proof. In order to show the first part, assumea∈F and letbbe any element of A. We will show thataia∨ib from which it follows by (F3) thata∨ib∈F.
This result is easily obtained, sincea∨i(a∨ib)∨i⊥= (a∨ia)∨ib∨i⊥=a∨ib∨i⊥.
In order to show the second part of the lemma, assume that a∨jb ∈ F. By
proposition 3.3.3 (vii) it follows thata∨jbiaanda∨jbib. Again, by (F3)
it follows that bothaandb are inF.
Proposition 3.4.5. If F is ani-prime filter, then{¬ija|a∈F} is aj-prime
filter.
Proof. Let F be an i-prime filter of a multi-Boolean algebra and let G =
{¬ija|a∈F}. We need to check thatGis a j-prime filter.
(F1) ¬ij⊤j =⊤i∈F by 3.3.3 (iii) thus⊤j∈G.
(F2) Assumea, b∈G, then¬ijaand¬ijbare in F and thus¬ija∨j¬ijb∈F.
It follows by (MBn7) that ¬ij(a∨ib) ∈ F and thus a∨i b ∈ G. For
k /∈ {i, j} we have¬ija∨k¬ijb∈F and by (MBn8)¬ij(a∨kb)∈F and
thusa∨kb∈G.
(F3) Assumea∈Gandaj b. We know that¬ija∈F. By 3.3.3(vi) it follows
that¬ijai¬ijb and thus we obtainb∈G.
(Pr) ⊥i ∈/ F. It follows by 3.3.3 (iii) that¬ij⊥j ∈/ F and thus⊥j ∈/ G. The
filterGis proper.
(F4) Supposea∨jb∈G, then¬ij(a∨jb)∈F and thus¬ija∨i¬ijb∈F from
which it follows by (F4) that ¬ija∈ F or ¬ijb ∈F. Thus either a or b
must be contained inG.
Definition 3.4.6. Given a multi-Boolean algebraA= (A,∨i,¬ij,⊥i)i,j∈A,a∈
Aandi∈Awe define the sets↓ia and↑iaas follows:
↓ia={b∈A|bia}
and,
↑ia={b∈A|aib}.
Given a setC⊆A the sets↓iC and↑iC are defined as
↓iC={b∈A|bic for somec∈C}
↑iC={b∈A|cib for somec∈C}.
These sets will play an important role in proving our desired representation theorem.
Proposition 3.4.7. For any multi-Boolean algebra A = (A,∨i,¬ij,⊥i)i,j∈A
anda∈A:
(i) The set↓ia is ani-ideal ofA.
(ii) The set ↑ia is ani-filter of A.
(iii) For anyi-idealI of A, the set↓i{a∨ic|c∈I} is an i-ideal.
(iv) For any i-filterF ofAandj ∈Asuch that j6=i the set↑i{a∨jb|b∈F}
is ani-filter.
Proof. Fix a multi-Boolean algebra A= (A,∨i,¬ij,⊥i)i,j∈A anda∈A. It will
turn out that in each of the four cases, showing the second clause (That is, (F2) in (ii) and (iv), and (I2) in (i) and (iii)) is the hardest part. For proofs of (i), (ii) and (iii) we refer to the appendix. Here, we will focus on proving item (iv), since it involves the most laborious effort.
LetF be an arbitraryi-filter ofAandj∈Asuch that j6=i. We will show that↑i{a∨jb|b∈F}is ani-filter. LetGto be the set
G=↑i{a∨jb|b∈F}
We need to show thatGsatisfies the properties (F1)-(F3). Proving (F1) and (F3) are relatively easy, whereas showing that (F2) is satisfied involves a fair amount of effort.
(F1) By assumption⊤i∈F. From this it follows thata∨j⊤i∈G. By (MB4)
it follows thata∨j⊤i≤i⊤i, hencea∨j⊤ii⊤i and thus⊤i ∈G.
(F2) Assumeeande′ are inG, we need to show thate∨
ke′∈Gfor anyk6=i.
Our strategy will be to find an elementdsuch that there is a b∈F such that a∨jb i d, and di e∨ke′. By transitivity of i it then follows
thata∨jbi e∨ke′ and hencee∨ke′∈G.
By assumption thate, e′ are inGit follows that there is are two elements
c, c′ contained inF such that a∨
j c i e and a∨j c′ i e′. In order to
improve readability of what there is to come, let’s denote v=a∨jc and
v′=a∨
jc′. Now, define das follows:
d= (v∨kv′)∨i(v∨ke′)∨i(e∨kv′).
We start by showing that there is a b ∈ F such that a∨j b i d. Our
Claim 3.4.8. a∨jbid
Proof. Since bothc andc′ are inF, it follows thatb=c∨
jc′ is inF as
well. Moreover, it follows that
a∨jb = a∨j(c∨jc′)
= (a∨jc)∨j(a∨jc′)
= v∨jv′
By proposition 3.3.3 we know that v∨j v′ =i v∨k v′ since i /∈ {j, k}.
Hence,a∨jb=i v∨kv′, from which we obtain thata∨jbi v∨kv′. Since
a∨jb ∈ G it follows that v∨k v′ is as well. For any xand y, we know
thatx≤i y, sincex∨ix∨iy=x∨iy. Hence we know thatxix∨iy as
well (by property 3.3.3 (viii)). It follows from this fact thatv∨kv′ id.
Hence,a∨jbid
Now that we have shown that there is absuch thata∨jbid(and hence
thatd∈G), it remains to be shown thatdie∨ke′.
Claim 3.4.9. die∨ke′
Proof. By definition ofiand by assumption thata∨jcieanda∨jc′ i
e′, we have the following equalities:
(a∨jc)∨ie∨i⊥ = v∨ie∨i⊥ = e∨i⊥,
(a∨jc′)∨ie′∨i⊥ = v′∨ie′∨i⊥ = e′∨i⊥.
From these equalities we obtain:
(e∨ke′)∨i⊥ = (e∨i⊥)∨k(e′∨i⊥) (3.2) = (v∨ie∨i⊥)∨k(v∨ie′∨i⊥) = (e∨kv′)∨i(v′∨ke′)∨i(v∨k⊥) ∨i(e∨kv′)∨i(e∨ke′)∨i(e∨k⊥) ∨i(⊥ ∨kv′)∨i(⊥ ∨ke′)∨i⊥ ∗ = (v∨kv′)∨i(v∨ke′)∨i(e∨kv′)∨i(e∨ke′)∨i⊥
Here the last equality (*) follows by application of (MB5): by reflexivity of
≤i, a formula of the form (⊥∨kv)∨i⊥is equivalent to (⊥∨kv)∨i⊥∨i⊥and
hence by (MB5) to⊥. It follows that we can eliminate all the ‘conjuncts’ (v∨k⊥) from
(v∨kv′)∨i(v∨ke′)∨i(v∨k⊥)
∨i(e∨kv′)∨i(e∨ke′)∨i(e∨k⊥)
∨i(⊥ ∨kv′)∨i(⊥ ∨ke′)∨i⊥
(v∨kv′)∨i(v∨ke′)∨i(e∨kv′)∨i(e∨ke′)∨i⊥.
We may conclude from (3.2) that (v∨kv′)∨i(v∨ke′)∨i(e∨kv′)i(e∨ke′)
and hence,di(e∨ke′).
To summarize, we have shown that there is an elementdand an element b∈F such thata∨jbi die∨ke′. Hence, e∨ke′∈Gas desired.
(F3) Letebe inGand assumeeif, then for somec∈F we havea∨jcie
and sinceeif it follows by transitivity thata∨jcif. Hence,f ∈G.
Note that, in general, thei-ideals andi-filters described in the above proposi- tion are not prime. Next we will prove a multi-player analogue of the prime-filter theorem.
Theorem 3.4.10. LetA= (A,∨i,¬ij,⊥i)i,j∈A be a multi-Boolean algebra. Let
J be an i-ideal and G an i-filter of A such that J ∩G=∅. Then there exists
I∈ Iip(A)and F=A\I∈ Fip(A)such that J ⊆I andG⊆F.
Proof. The proof of this theorem will be structurally similar to the proof of Theorem 9.13 in [4].
Define
E={K∈ I(A)|J⊆K andK∩G=∅}
We will show that (E,⊆) has a maximal element I. First of all, E containsJ and so is nonempty. LetC ={Kα|α∈λ} be a chain in E. We need to show
thatK=Sα∈λKα∈ E. CertainlyJ⊆KandK∩G=∅(for if not, then there
would be some α∈ λ such that Kα∩G6= ∅). What remains to be shown is
thatK∈ I(A).
Claim 3.4.11. K is an i-filter
Proof. (I1) follows is immediately since ⊥i ∈Kα for each Kα ⊆K. Also (I3)
is easy to obtain since if a∈ K, then a∈ Kα for some α∈ λ. If in addition
and a i b, thenb ∈ Kα as well and hence b ∈ K. As for (I2), assume that
a, b∈K. We need to show thata∨ib∈K. By assumption, there areα, β∈λ
such thata∈Kα andb∈Kβ. SinceC is a chain, we may assume, without loss
of generality thatKα⊆Kβ. Thus it follows thatb∈Kβ and alsoa∨ib∈Kβ,
from which we conclude thata∨ib∈K. It follows that K is ani-filter.
Hence,K∈ E. By Zorn’s lemma we may now conclude thatEhas a maximal elementI. ThisIwill be our candidate to prove the theorem.
Proof. We know know that I is ani-ideal since it is inE. The only thing that there is left to be shown is that it is an i-prime ideal. Clearly, I is proper for we know that G is nonempty. (I4) Suppose that for some j 6= i ∈ A we havea∨jb∈I buta, b /∈I. Because I is assumed to be the maximal element
of E, any ideal properly containing I intersects with G. Therefore the i-ideal Ia =↓i{a∨ic|c∈I}intersects withG. From this we may conclude that there
exists aca∈I such thatgia∨ica for someg∈G. SinceGis assumed to be
ani-filter, we know thata∨ica∈G. By similar reasoning we can find acb∈I
such thata∨icb∈G. The following equivalence is valid inA:
(a∨jb)∨i(ca∨icb) = ((a∨ica)∨icb)∨j((b∨icb)∨ica) (3.3)
SinceG is a filter, and sincea∨ica is inG, it follows by lemma 3.4.4 that
((a∨ica)∨icb)∈G. Similarly, we know that ((b∨icb)∨ica)∈G. By (F2) it
follows that ((a∨ica)∨icb)∨j((b∨icb)∨ica) is inG. However, the left-hand
side of the equation 3.3 is inI. This is because bothcaandcbare inI, and thus
(ca∨icb)∈I. Since (a∨jb) was assumed to be inIwe have (a∨jb)∨i(ca∨icb)
inI. This contradicts the fact thatI∩G=∅. Thus we obtain thatI satisfies (I4), hence it is ai-prime ideal.
We may now conclude, (by proposition 3.4.3), that F = A\I an i-prime filter.
Corollary 3.4.13 (i-Prime Filter Theorem). Given a multi-Boolean algebra
A = (A,∨i,¬ij,⊥i)i,j∈A, an i-filter G ⊂ A and an element a ∈ A such that
a /∈G, there is ai-prime filterF such thatG⊆F andF does not contain a. Proof. Consider the setI=↓ia. By proposition 3.4.7,Iis ani-ideal. Moreover,
I∩G = ∅. Suppose otherwise, then there is a b such that b ∈ I and b ∈ G, hence by definition of I, b i a. From (F3) it thus follows that a∈ Gwhich
contradicts our assumption thata /∈G. By theorem 3.4.10 it follows that there is ani-prime filterF extendingGsuch thata /∈F.
Theorem 3.4.14 (Representation Theorem). Every multi-Boolean algebra is isomorphic to a concrete multi-player algebra.
Proof. Fix a multi-Boolean algebra A = (A,∨i,¬ij,⊥i)i,j∈A. We will fix one
player, player 0 ∈ A and embed A into the concrete multi algebra C(F0p(A)).
Remember that F0p(A) is the collection of 0-prime filters of A. Consider the map: ρ:A→C(F0p(A)) defined as follows:
ρa(F)(j) =
wif¬0ja∈F
lotherwise
We need to show thatρis an injective homomorphism. First we will show that ρis a homomorphism. Because the proof of this fact involves some convoluted arguments combining many of the propositions proved in this section, we will go through the proof in great detail.
Let us first consider the ∨l operator. We need to show that ρ(a∨lb) =
ρ(a)∪lρ(b). This boils down to showing that
ρa∨lb(F)(j) =wiff (ρa∪lρb)(F)(j) =w.
We distinguish two cases, l= 0 andl 6= 0. That is, we distinguish between
∨0and∨l withl6= 0. Case∨0. Forj6= 0, ρa∨0b(F)(j) =w ⇔ ¬0j(a∨0b)∈F ⇔ ¬0ja∨j¬0jb∈F ∗ ⇔ ¬0ja∈F and ¬0jb∈F ⇔ ρa(F)(j) =w andρb(F)(j) =w ⇔ (ρa∪0ρb)(F)(j) =w
Note that the crucial step is the equality (*). The direction from left to right follows from 3.3.3 (vii) and the other direction follows by clause (F2) of the definition of 0-filter. The equality of the last two lines follows from the definition of concrete multi-player set algebras in section 3.2.
Forj= 0, ρa∨0b(F)(0) =w ⇔ ¬00(a∨0b)∈F ⇔ ¬00a∨0¬00b∈F ∗ ⇔ ¬00a∈F or¬00b∈F ⇔ ρa(F)(0) =wor ρb(F)(0) =w ⇔ (ρa∪0ρb)(F)(0) =w
Again, the crucial step is the (*)-equality. Here, the direction from left to right follows from the assumption thatF is a 0-prime filter. The other direc- tion follows from 3.4.4. Again, the equality of the last two lines follows from the definition of concrete multi-agent powerset algebras in section 3.2.
Case∨l, withl6= 0.
ρ(a∨lb)(F)(j) =w ⇔ ¬0j(a∨lb)∈F ⇔ ¬0ja∨l¬0jb∈F ⇔ ¬0ja∈F and¬0jb∈F ⇔ ρa(F)(j) =wandρb(F)(j) =w ⇔ (ρa∪lρb)(F)(j) =w and forj=l, ρ(a∨lb)(F)(l) =w ⇔ ¬0l(a∨lb)∈F ⇔ ¬0la∨0¬0lb∈F ⇔ ¬0la∈F or¬0lb∈F ⇔ ρa(F)(l) =wor ρb(F)(l) =w ⇔ (ρa∪lρb)(F)(l) =w
Now, let us have a look at the other connective, ¬ij.We will show that
ρ(¬ija) =∼ijρ(a)
This boils down to showing that
ρ¬ija(F)(k) =w iff∼ij ρa(F)(k) =w.
Again, we will consider two cases 0∈ {i, j} and 0∈ {/ i, j}. If 0∈ {i, j}, we can need to distinguish between two cases¬0j and¬i0. Here, we will only treat
the first case and leave the second to the reader. Thus, here we will discuss the following two negations: ¬0j withj can be equal to 0 or not, and¬ij with both
iandj distinct from 0.
Case¬0j. Fork= 0, ρ¬0ja(F)(0) =w ⇔ ¬00¬0ja∈F ⇔ ¬0ja∈F ⇔ ρa(F)(j) =w ⇔ ∼0jρa(F)(0) =w
Here, the equivalence between the first and the second line follows from (MBn1).
ρ¬0ja(F)(j) =w ⇔ ¬0j¬0ja∈F
⇔ ¬0j¬j0a∈F
⇔ ¬00a∈F
⇔ ρa(F)(0) =w
⇔ ∼0jρa(F)(j) =w
Here, the equivalence between the second and third line follows from (MBn3). Fork /∈ {0, j},
ρ¬0ja(F)(k) =w ⇔ ¬0k¬0ja∈F
⇔ ¬0ka∈F
⇔ ρa(F)(k) =w
⇔ ∼0jρa(F)(k) =w
Here the equivalence between the first and the second line follows from (MBn4) and the observation 3.1. The last step follows from the definition of∼
in the concrete multi-player algebra.
Case¬ij, with 0∈ {/ i, j}. Fork=i, ρ¬ija(F)(i) =w ⇔ ¬0i¬ija∈F ⇔ ¬0ja∈F ⇔ ρa(F)(j) =w ⇔ ∼ij ρa(F)(i) =w and fork /∈ {i, j}, ρ¬ija(F)(k) =w ⇔ ¬0k¬ija∈F ⇔ ¬ij¬0ka∈F ⇔ ¬0ka∈F ⇔ ρa(F)(k) =w ⇔ ∼ij ρa(F)(k) =w
Here the equivalence between the first and second line follows from (MBn6)