Download Subalgebras of the free Heyting algebra on one generator

Subalgebras of the free Heyting algebra on one generator
R. Elageili
Department of Mathematics, Benghazi University,
Benghazi, Libya, e-mail [email protected]
Abstract
In this paper we describe the subalgebras of the free Heyting algebra A1 on one generator,
generated by an arbitrary single element of A1 and we give a general theorem which provides
an explicit classification of all the subalgebras of A1 .
1
Introduction
Heyting algebras are a generalization of Boolean algebras; the most typical example is the lattice
of open sets of a topological space. It is well known that Heyting algebras are algebraic models of
intuitionistic propositional logic, which is properly contained in classical propositional logic.
Heyting algebras are special distributive lattices, and they form a variety.
Definition: An algebra A = (A, ∨, ∧, →, 0, 1) is a Heyting algebra if it is a bounded distributive
lattice with least element 0 and greatest element 1, and for all x, y ∈ A, x → y is the greatest
element z of A such that x ∧ z ≤ y, (where ≤ is defined by: x ≤ y if and only if x ∧ y = x). This
element x → y is called the pseudo-complement of x with respect to y.
In any Heyting algebra the pseudo-complement ¬x of x is defined by ¬x = x → 0. Note that
x ∧ ¬x = 0, and that ¬x is the greatest element having this property.
A complete Heyting algebra is a Heyting algebra that is a complete lattice (every subset has a
supremum).
Let V be a variety. Recall that an algebra A ∈ V is said to be a free algebra over V , if there
exists a set E ⊆ A such that E generates A and every mapping from E to an algebra B ∈ V can
be extended uniquely to a homomorphism from A to B. In this case E is said to be a set of free
generators of A. If E is finite then A is said to be a finitely generated free algebra.
We denote a finitely generated free Heyting algebra with α free generators (which is uniquely
determined up to isomorphism) by Aα . Free Heyting algebras arise naturally as the the Lindenbaum
algebras of intuitionistic propositional logic (IPC) with α propositional variables over the empty
theory.
In contrast with Boolean algebras, finitely generated free Heyting algebras are infinite as was
shown by Mckinsey and Tarski in the 1940s ([7]). For one generator, A1 is well understood (The
Rieger Nishimura ladder, ([4], [5])), but from two on, the structure remains mysterious, though many
properties are known.
With the help of recursively described Kripke models Hα , Bellissima ([1]) gave a representation
of Aα . Essentially the same construction is due to Grigolia ([3]) and Esakia ([6]).
The free Heyting algebra A1 on one generators may be defined as the Lindenbaum algebra of
intuitionistic propositional logic IP C on a set P = P1 of one propositional variable, this is the
so-called ‘Rieger–Nishimura lattice’, or ‘ladder’, [5, 4], has an explicit description as a lattice; all
elements which are not its least or greatest elements 0, 1, lie in antichains of size 2, of which there
1 keywords:
free Heyting algebra, subalgebra
1
are countably many arranged in ω levels, and the partial order relation between these is quite easily
and explicitly described.
Our aim in this paper is to provide an explicit classification of all the subalgebras of A1 ..
Definition: Let A be an algebra, X ⊆ A. The subalgebra generated by X is the intersection of all
algebras containing X written hXi. (This can be constructed by closing up X under the operations.)
The Rieger-Nishimura Ladder, A1 , is shown in figure below.
1
•
•
¬¬p −→ (p ∨ ¬p)
¬p ∨ ¬¬p
p ∨ ¬p
¬¬p
¬p
p
0
2
Subalgebras of A1
First I give the subalgebras of A1 which are generated by only one element.
Example 2.1. S is the subalgebra generated by either 0 or 1.
Then S contains only {0, 1} see figure 1.
1
0
Figure 1:
Example 2.2. S is the subalgebra generated by p.
Then S equals A1 see figure ??.
Example 2.3. The set {0, ¬p, ¬¬p, ¬p ∨ ¬¬p, 1}, as in figure 2 is equal to the subalgebra S generated
by either ¬p or ¬¬p.
2
→
0
¬p
¬¬p
¬p ∨ ¬¬p
1
0
1
¬¬p
¬p
0
0
¬p
1
1
¬p
¬p
¬p
¬¬p
1
¬¬p
1
¬¬p
¬¬p
¬p ∨ ¬¬p
1
1
1
1
¬p ∨ ¬¬p
1
1
1
1
1
1
Table 1:
Indeed, it is closed under {¬, ∧, ∨, →}, since ¬¬¬p = ¬p, ¬p ∧ ¬¬p = 0, ¬0 = 1 and Table 1
shows it is closed under →.
1
¬p ∨ ¬¬p
E
zz
zz
z
zz
zz
z
z
¬p D
DD
DD
DD
DD
DD
D
0
EE
EE
EE
EE
EE
_E _ _ ¬¬p _x _ _
x
xx
x
xx
xx
x
xx
Figure 2:
Theorem 2.4. : If x ∈ Aα , and x is above every atom, then ¬x = 0.
Proof: Suppose that ¬x 6= 0. Then ¬x ≥ y for some atom y. But x ≥ y, hence x ∧ ¬x ≥ y which
is a contradiction. 2
Hence the negation of any element of x ∈ A1 apart from 0, p, ¬p, ¬¬p is equal to 0.
Example 2.5. S is the subalgebra generated by such x.
Then S is the chain {0, x, 1} see figure 3.
1
x
0
Figure 3:
The previous examples gave a complete description of the subalgebras generated by arbitrary
single element of A1 , and it follows from them that subalgebras of free algebras need not be free.
3
Now, in general I will use the following notation for the elements of A1 : 0 = ⊥, a1 = ¬p, b1 = p,
an+1 = an → bn , bn+1 = an ∨ bn , and 1 = p → p, see figure ??.
Note that ¬a1 = a2 , ¬a2 = a1 , and ¬ai = 0 for all i ≥ 3, ¬bi = 0 for all i ≥ 2.
1
b4
a4
a3
b
w 3
ww
w
ww
ww
w
w
ww
ww
a2
b2 GG
GG
GG
GG
GG
GG
GG
a15
55
55
55
55
0
b1
Figure 4:
Lemma 2.6. : The subalgebra S generated by each of the following sets:
1. {a2 , b2 }
2. {a1 , b2 }
3. {a1 , a3 }
4. {a2 , a3 }
is equal to A1 .
Proof: To prove this lemma it is enough to generate b1 inside S.
1. a2 ∧ b2 = b1
2. ¬a1 = a2 and a2 ∧ b2 = b1
3. ¬a1 = a2 , a1 ∨ a2 = b3 , a3 ∧ b3 = b2 , and a2 ∧ b2 = b1
4. a2 ∧ a3 = b1 . 2
Definition: We say that a subset X of A1 is nearly A1 if its complement is finite. A partially ordered
set (X, <) is quasi-linearly ordered if there is a linearly ordered set (Y, <) and a homomorphism θ
from X onto Y such that for each y, θ−1 (y) is either a singleton or a diamond.
Lemma 2.7. : The subalgebra S generated by the set {bn , bn+1 } for n ≥ 2 is nearly A1 .
Proof: As bn is above all the atoms, ¬bn = 0. Also bn → bn+1 = 1, bn+1 → bn = an+1 , an+1 ∨bn+1 =
bn+2 , and an+1 → bn+1 = an+2 , and so on, so we get all the points of A1 above bn as in figure 5. 2
4
1
·
·
_ _ _ an+2 _ _ _
bn+2 L
an+1:
LL
LL
LL
LL
LL
LL
L
bn+1
::
::
::
::
:
bn
0
Figure 5:
Corollary 2.8. : The subalgebra S generated by the set {an , bn } for n ≥ 3 is nearly A1 .
Proof: This is an immediate consequence of the previous lemma, since an ∧ bn = bn−1 . 2
Lemma 2.9. : The subalgebra S generated by the set {an+1 , bn } (or the set {an+2 , bn }) for n ≥ 2
is quasi-linearly ordered, with just one diamond.
Proof: As bn is above all the atoms, ¬bn = 0. Also an+1 → bn = an+2 , bn → an+1 = 1, and
an+2 → bn = an+1 , bn → an+2 = 1, and an+1 → an+2 = an+2 , an+2 → an+1 = an+1 , and
an+1 ∨ an+2 = bn+3 . Hence we get 0, bn , an+1 , an+2 , bn+3 , 1 which is a diamond starting at bn , see
figure 6. To verify that it is a subalgebra, note that it is clearly a lattice, and it is also closed under
→ as shown in Table 2. 2
→
0
bn
an+1
an+2
bn+3
1
0
1
0
0
0
0
0
bn
1
1
an+2
an+1
bn
bn
an+1
1
1
1
an+1
an+1
an+1
Table 2:
5
an+2
1
1
an+2
1
an+2
an+2
bn+3
1
1
1
1
1
bn+3
1
1
1
1
1
1
1
1
bn+3>
>>
>>
>>
>>
>
_ _ _ an+1> _ _ _>
>>
>>
>>
>>
an+2
bn
0
Figure 6:
Corollary 2.10. : The subalgebra S generated by the set {an , an+1 } for n ≥ 3 is quasi-linearly
ordered.
Proof: Since an ∧ an+1 = bn−1 . 2
Theorem 2.11. : The subalgebras of A1 are all of one of the following forms:
1. Nearly A1 (differs from A1 on a finite set)
2. Quasi-linearly ordered (linearly ordered and decorated by diamonds).
Proof: Let S be a subalgebra of A1 .
By Example 2.2 and Lemma 2.6 we may assume that S does not contain any of the sets
{b1 }, {a2 , b2 }, {a1 , b2 }, {a1 , a3 } or {a2 , a3 } since then we obtain A1 itself.
If S contains {bn , bn+1 } for some n ≥ 2, then we use Lemma 2.7 to see that S contains a subalgebra
which is nearly A1 , so is nearly A1 itself.
Now suppose that S does not contain any of these subsets. Let x and y be any incomparable
members of S. Then by inspection, {x, y} = {an , bn } or {an , an+1 } for some n. Corollary 2.8 tells
us about the first, and Corollary 2.10 about the second.
Let D1 , D2 be any two diamonds occurring in S. Then D1 = {bm , am+1 , am+2 , bm+3 } and D2 =
{bn , an+1 , an+2 , bn+3 } for some m, n. Suppose that m < n. Then n 6= m + 1 or m + 2 as the
former gives {bm , bm+1 } ⊆ S and the latter gives {bn , bn+1 } ⊆ S, contrary to assumption. Therefore
m + 3 ≤ n and D1 ≤ D2 . Since all points of S not in diamonds are comparable with all members of
S it follows that S is quasi-linearly ordered. 2
Now using Theorem 2.11 we can explicitly describe all subalgebras of A1 . If S is nearly A1 then
the least x ∈ S such that for all y ≥ x, y ∈ S must equal bn for some n (where this includes the
possibility of n = 0 where bn = 0) and then since an ∧ bn = bn−1 , for n > 0 we have an ∈
/ S. Using
the the proof of Theorem 2.11 we can then see that the {y ∈ S : y ≤ x} is quasi-linearly ordered.
Which quasi-linearly ordered sets can arise is restricted in a similar way to what follows.
To complete our description we must just characterize which S arise as quasi-linearly ordered.
Let X, Y and Z be three disjoint subsets of {n : n ≥ 2} satisfying:
1. If m < n in X then n = m + 3 or n > m + 4.
2. If n ∈ X, m ∈ Y then n 6= m ± 1 and n + 3 6= m ± 1.
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3. If n ∈ X, m ∈ Z then m 6= n ± 1 and m 6= n ± 2.
4. If m < n in Y then n 6= m + 1.
5. If m < n in Z then n 6= m + 1.
6. If m < n such that m ∈ Y and n ∈ Z then n 6= m + 1 and n 6= m + 2.
7. If m < n such that m ∈ Z and n ∈ Y then m > 2.
To see where these clauses come from, suppose that S is quasi-linearly ordered, and let
X = {n : bn is the least point of a diamond of S},
Y = {n ∈
/ X ∪ (X + 3) : bn ∈ S}, and
Z = {n ∈
/ (X + 1) ∪ (X + 2) : an ∈ S}, where X + i = {n + i : n ∈ X}.
Thus the diamonds of S are {bn , an+1 , an+2 , bn+3 } for n ∈ X, the isolated bn s of S are bn for
n ∈ Y and the isolated an s in S are an for n ∈ Z. Then the seven clauses are true as follows.
Clause 1 says that two diamonds do not overlap, though they can ‘touch’, but also they cannot be
exactly 1 apart because of Lemma 2.7.
Clause 2 says that there are no two consecutive bn s (again because of Lemma 2.7).
Clause 3 similarly deals with bn related to an±1 , an±2 (using Lemma 2.9).
Clause 4 says that two isolated bn s cannot be consecutive (Lemma 2.7).
Clause 5 says that two isolated an s cannot be consecutive (Lemma 2.10).
Clause 6 corresponds to Lemma 2.9.
Clause 7 corresponds to the fact that ¬a2 = a1 .
The explanations just given show that if S is quasi-linearly ordered then X, Y and Z must filfil
all these clauses. Conversely, suppose that all clauses are filfilled. Then S is quasi-linearly ordered
and is a subalgebra. It is obviously a sublattice, so we just need to show that it is closed under →.
Now we have already seen that each diamond is closed under →. Also if {bm , am+1 , am+2 , bm+3 }∪
{bn } ⊆ S where 2 ≤ m < n − 3 then by the first clause m < n − 4 and bn → bm = bm , bn →
am+1 = am+1 , bn → am+2 = am+2 , and bn → bm+3 = bm+3 , if {bm } ∪ {bn , an+1 , an+2 , bn+3 } ⊆ S
where 2 ≤ m < n then by the first clause m < n − 1 and bn → bm = bm , an+1 → bm = bm ,
an+2 → bm = bm , and bn+3 → bm = bm .
If {bm , am+1 , am+2 , bm+3 } ∪ {an } ⊆ S where 2 ≤ m < n − 3 then by the third clause m < n − 5
and an → bm = bm , an → am+1 = am+1 , an → am+2 = am+2 , and an → bm+3 = bm+3 , if
{am } ∪ {bn , an+1 , an+2 , bn+3 } ⊆ S where 2 < m < n then bn → am = am , an+1 → am = am ,
an+2 → am = am , and bn+3 → am = am .
Finally if {bm } ∪ {an } ⊆ S where 2 ≤ m < n then by clause 6, m < n − 2 and an → bm = bm ,
and if {am } ∪ {bn } ⊆ S where 2 < m < n then bn → am = am .
Remark: There are only countably many subalgebras which are nearly A1 since there are only
countably many finite sets which can be omitted, and there are uncountably many subalgebras
which are quasi-linearly ordered since if S is quasi-linearly ordered and infinite then we can omit any
set of isolated points or diamonds in 2ℵ0 ways and obtain other quasi-linearly ordered subalgebras.
References
[1] F. Bellissima, Finitely generated free Heyting algebras, Journal of Symbolic Logic, 51 (1986),
152-165.
[2] R. Elageili, Free Heyting algebras, PhD thesis, University of Leeds, 2011.
[3] R. Grigolia, Free and projective Heyting and monadic Heyting algebras, in: Höhle, U., Klement,
E.P. (eds) Non-Classical Logics and Their Applications to Fuzzy Subsets, Kluwer Academic
Publisher, Dordrecht (1995) Kluwer Acad. Publishers, 33-52.
[4] I. Nishimura, On formulas in one variable in intuitionistic propositional calculus, Journal of
Symbolic Logic, 25 (1960), 327-331.
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[5] L. Riger, Zametki o t. naz. svobodnyh algebrah c zamykaniami, Czechoslovak Mathematical
Journal 7 (82) (1957), 16-20.
[6] L. Esakia, Topological Kripka model, Soviet Math. Dokl.15 (1974), 147-151.
[7] J. Mckinsey, and A. Tarski, On closed elements in closure algebras, Annals of Mathrmatics.47
(1946), 122-162.
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