Ernesto Rude Jauregui
REGÍMENES
3.4 El aspecto político
Definition 3.6.1 We say that a bounded pseudo-BCK-semilattice (A, ∨, →, , 0, 1) is (i) involutive if it satisfies the identities x = x−∼= x∼−,
(ii) good if it satisfies the identity x−∼= x∼−.
For instance, every pseudo-MV-algebra is an involutive pseudo-BCK-semilattice, and every bounded BCK-semilattice is a good pseudo-BCK-semilattice.
If (A, ∨, →, , 0, 1) is involutive then the maps a 7→ a− and a 7→ a∼ are mutually
inverse antiautomorphisms of the poset (A, ≤), which is therefore a lattice where the meet a ∧ b is given by
a ∧ b = (a−∨ b−)∼= (a∼∨ b∼)−.
In Section 3.4 we have defined the addition ⊕ the properties of which turn out to be quite nice in case of good pseudo-BCK-semilattices. Recall that
a ⊕ b = a∼→ b−∼= b− a∼−.
Lemma 3.6.2 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice. Then: (1) a ⊕ 0 = 0 ⊕ a = a−∼= a∼−, (2) a ⊕ 1 = 1 ⊕ a = 1, (3) (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c), (4) if a ≤ b then a ⊕ c ≤ b ⊕ c and c ⊕ a ≤ c ⊕ b, (5) a ∨ b ≤ a ⊕ b, (6) a ⊕ b = (a ⊕ b)−∼= a−∼⊕ b−∼.
Proof: As a sample we show the property (3): (a ⊕ b) ⊕ c = c− (a∼→ b−∼)∼−=
c− (a∼→ b−∼) = a∼→ (c− b−∼) = a∼→ (c− b−∼)−∼= a ⊕ (b ⊕ c).
By a bounded dually integral dually residuated lattice we mean an algebra (A, ∨, ∧, ⊕, , ;, 0, 1) where (A, ∨, ∧, 0, 1) is a bounded lattice, (A, ⊕, 0) is a monoid, and
a ⊕ b ≥ c iff a ≥ c b iff b ≥ c ; a for all a, b, c ∈ A (see Example 1.1.6).
Proposition 3.6.3 For (A, ∨, →, , 0, 1) an involutive pseudo-BCK-semilattice, (1) (A, ∨, ∧, ⊕, , ;, 0, 1), where x y = (x y)− and x ; y = (x → y)∼, is a
bounded dually integral dually residuated lattice,
(2) (A, ∨, ∧, ·, →, , 0, 1), where x · y = (x−⊕ y−)∼, is a bounded integral residuated
3.6. Good pseudo-BCK-semilattices 61
Proof: (1) We know that (A, ∨, ∧) is a bounded lattice, due to the previous lemma, (A, ⊕, 0) is a monoid and we also have z ≤ x ⊕ y = x∼ → y iff x∼ ≤ z y iff
x ≥ (z y)−= z y, and similarly, z ≤ x ⊕ y = y− x iff y ≥ (z → x)∼ = z ; x.
(2) Note that z ≥ (x → y−)∼ iff z− ≤ x → y− iff x ≤ z− y− = y → z iff
y ≤ x z = z∼→ x∼ iff z∼≤ y x∼ iff z ≥ (y x∼)−, whence it follows
x · y = (x−⊕ y−)∼= (x → y−)∼= (y x∼)−= (x∼⊕ y∼)−.
This entails that (A, ·, 1) is a monoid, and x · y ≤ z iff x ≤ y → z iff y ≤ x z. Given a good pseudo-BCK-semilattice (A, ∨, →, , 0, 1), extending the analogy with pseudo-MV-algebras, we introduce the partial addition + as a restriction of ⊕:
a+b is defined iff a−∼≤ b−(iff a∼ ≥ b−∼), in which case we put a+b = a⊕b.
The following definition is motivated by Georgescu’s definition of Rieˇcan states on good pseudo-BL-algebras (cf. [25]).
Definition 3.6.4 We call a mapping s : A → [0, 1] a Rieˇcan state (or an additive measure) on a good pseudo-BCK-semilattice (A, ∨, →, , 0, 1) if
(i) s(1) = 1,
(ii) s(a + b) = s(a) + s(b) provided a + b is defined in A.
Lemma 3.6.5 Let s be a Rieˇcan state on a good pseudo-BCK-semilattice. Then (1) s(a−) = s(a∼) = 1 − s(a),
(2) s(a−∼) = s(a∼−) = s(a),
(3) s(0) = 0,
(4) b ≤ a implies s(b) ≤ s(a) and s(a → b−∼) = s(a b∼−) = 1 − s(a) + s(b),
(5) s(a → b−∼) = s(a b∼−) = 1 − s(a ∨ b) + s(b).
Proof: (1) For any a ∈ A, a−+ a is defined (since a−∼≥ a−∼) and equals a−⊕ a =
a−∼ → a−∼ = 1. Hence we have 1 = s(1) = s(a−) + s(a), so s(a−) = 1 − s(a).
Analogously, a + a∼= 1 and s(a∼) = 1 − s(a).
(2) This follows immediately from (1).
(3) For every a ∈ A, a + 0 exists and is equal to a−∼, hence s(a) = s(a−∼) = s(a + 0) = s(a) + s(0) and so s(0) = 0.
(4) Both a−+ b and b + a∼ exist and we have a−⊕ b = b− a−∼−= b− a−=
a → b−∼ and similarly b ⊕ a∼ = a b∼−. Therefore s(a → b−∼) = s(a−) + s(b) =
1 − s(a) + s(b) and s(a b∼−) = s(b) + s(a∼) = s(b) + 1 − s(a), which also proves
s(b) ≤ s(a).
(5) We have (a ∨ b) → b−∼ = a → b−∼ and (a ∨ b) b∼− = a b∼− which
due to (4) yields s(a → b−∼) = s((a ∨ b) → b−∼) = 1 − s(a ∨ b) + s(b) and likewise
Proposition 3.6.6 Every Bosbach state on a good pseudo-BCK-semilattice A is a Rieˇcan state. If, moreover, A is involutive then every Rieˇcan state is a Bosbach state.
Proof: Let s be a Bosbach state on A. Assume that a + b is defined, i.e., a∼≥ b−∼.
Then a + b = a ⊕ b = a∼ → b−∼ and s(a + b) = s(a∼→ b−∼) = 1 − s(a∼) + s(b−∼) = 1 − (1 − s(a)) + s(b) = s(a) + s(b). Since s(1) = 1, s is a Rieˇcan state.
Assume now that A is an involutive pseudo-BCK-semilattice. Then, by Lemma 3.6.5 (3), (5), every Rieˇcan state s satisfies s(0) = 0 and s(a → b) = 1 − s(a ∨ b) + s(b), so s is a Bosbach state.
In contrast to good pseudo-BL-algebras or good R`-monoids, we can merely say that Bosbach states are a subclass of Rieˇcan states, but the converse fails to be true, that is, a Rieˇcan state need not be a Bosbach one:
Example 3.6.7 Consider the following BCK-semilattice (A, ∨, →, 0, 1):
→ 0 a b 1
0 1 1 1 1
a b 1 1 1
b b b 1 1
1 0 a b 1
This is a good pseudo-BCK-semilattice with 0 < a < b < 1. We have
⊕ 0 a b 1 0 0 b b 1 a b 1 1 1 b b 1 1 1 1 1 1 1 1 + 0 a b 1 0 0 b b 1 a b 1 1 · b b 1 1 · 1 1 · · ·
Hence the mapping s : A → [0, 1] defined by s(0) = 0, s(a) = s(b) = 12 and s(1) = 1 is a Rieˇcan state, but it is not a Bosbach state as s(a) + s(a → b) = s(a) + s(1) = 1
2 + 1 6= 12+ 12 = s(b) + s(b) = s(b) + s(b → a). An easy inspection shows that there is no Bosbach state on A (since {1} is the only proper deductive system).
For a good pseudo-BCK-semilattice (A, ∨, →, , 0, 1) we define
Reg(A) = {a ∈ A : a−∼= a} and H(A) = {a ∈ A : a−∼= 1}.
Let us call the elements of Reg(A) regular and those of H(A) dense. One readily sees that
Reg(A) = {a−: a ∈ A} = {a∼: a ∈ A} and
H(A) = {a ∈ A : a−= 0} = {a ∈ A : a∼= 0}.
Lemma 3.6.8 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice. Then Reg(A) a non-empty subalgebra of (A, →, , 0, 1) and H(A) is a proper deductive system.
3.6. Good pseudo-BCK-semilattices 63
Proof: Obviously, 0, 1 ∈ Reg(A). Assume a, b ∈ Reg(A). Then a → b = a → b∼−= (a → b∼−)∼− = (a → b)∼− and a b = a b−∼ = (a b−∼)−∼ = (a b)−∼, so
a → b, a b ∈ Reg(A).
We have 1 ∈ H(A) and 0 /∈ H(A). If a ∈ H(A) and a → b ∈ H(A), then 1 = (a → b)−∼ ≤ (a → b−∼)−∼ = a → b−∼, so that a → b−∼ = 1. But a → b−∼ =
a−∼→ b−∼= 1 → b−∼= b−∼, and hence b ∈ H(A).
In general, Reg(A) is not closed under ∨. On the other hand, for every a, b ∈ Reg(A) the infimum a ∧ b of {a, b} in A exists and is given by
a ∧ b = (a−∨ b−)∼= (a∼∨ b∼)−.
It is also obvious that a ∧ b ∈ Reg(A). The proof of the next result is straightforward. Proposition 3.6.9 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice. Then Reg(A), equipped with the restriction to Reg(A) of the underlying order ≤ of A, forms a lattice (Reg(A), t, ∧) in which
a t b = (a ∨ b)−∼ and a ∧ b = (a−∨ b−)∼= (a∼∨ b∼)−. Moreover, for all a, b ∈ Reg(A),
(a−t b−)∼= (a∼t b∼)−= a ∧ b and (a−∧ b−)∼= (a∼∧ b∼)− = a t b. Corollary 3.6.10 (Reg(A), t, →, , 0, 1) is an involutive pseudo-BCK-semilattice. Proposition 3.6.11 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice. Then the mapping
f : a 7→ a−∼
is a homomorphism of (A, ∨, →, , 0, 1) onto (Reg(A), t, →, , 0, 1) if and only if A satisfies the identities
(x → y)−∼ = x → y−∼,
(x y)−∼ = x y−∼. (3.6.1)
If this is the case, then H(A) ∈ DSc(A) and Reg(A) ∼= A/H(A).
Proof: For every a, b ∈ A, f (a ∨ b) = (a ∨ b)−∼= (a−∧ b−)∼ = (a−∼−∧ b−∼−)∼=
a−∼t b−∼ = f (a) t f (b), thus f preserves finite joins. It is plain that the equations
(3.6.1) give a necessary and sufficient condition for f to be a homomorphism, because f (a → b) = (a → b)−∼ and f (a) → f (b) = a−∼ → b−∼ = a → b−∼, and likewise
f (a b) = (a b)−∼ and f (a) f (b) = a b−∼. For the final claim, Ker(f ) =
{a ∈ A : f (a) = 1} = H(A) and therefore Reg(A) ∼= A/Ker(f ) = A/H(A).
Proposition 3.6.12 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice satisfying the identities (3.6.1). Then Bosbach and Rieˇcan states coincide.
Proof: Assume that s is a Rieˇcan state on A. Clearly, s(0) = 0. Let a, b ∈ A, a ≥ b. Then s(a → b) = s((a → b)−∼) = s(a → b−∼) = 1 − s(a) + s(b) by Lemma 3.6.5.
Remark 3.6.13 In [19], Dvureˇcenskij and Rach˚unek proved that Rieˇcan states on good R`-monoids coincide with Bosbach states. The proof consists in showing that every extremal Rieˇcan state is an extremal Bosbach state, and then applying the Krein- Mil’man Theorem (see our Corollary 3.5.2). The result can be achieved as a consequence of the previous proposition:
Let (L, ∨, ∧, ·, →, , 0, 1) be a good R`-monoid and s be a Rieˇcan state on L. By [62] (also [20, 19]), L satisfies the identities
(x → y)−∼= x−∼→ y−∼, (x y)−∼= x−∼ y−∼.
Since x−∼ → y−∼ = x → y−∼ and x−∼ y−∼= x y−∼, L satisfies the equations
(3.6.1), and hence every Rieˇcan state on L is a Bosbach state.
For both Bosbach and Rieˇcan states we have s(a−∼) = s(a). It comes therefore
as no surprise that s Reg(A), the restriction to Reg(A) of a state s (of either kind), is a state on Reg(A), from which s can be recovered by s(a) = s Reg(A) (a−∼), a ∈ A.
Since Reg(A) is an involutive pseudo-BCK-semilattice, it follows that Bosbach and Rieˇcan states on Reg(A) are identical, so the question arises which kind of states on A corresponds to states on Reg(A).
Proposition 3.6.14 Let (A, ∨, →, , 0, 1) be a good pseudo-BCK-semilattice and let s : Reg(A) → [0, 1] be a state on (Reg(A), t, →, , 0, 1). Then the map ˆs : A → [0, 1] defined by
ˆ
s(a) = s(a−∼)
is a Rieˇcan state on (A, ∨, →, , 0, 1) such that ˆs Reg(A)= s.
Proof: It is obvious that ˆs(1) = 1. Take a, b ∈ A such that a + b exists, i.e., a−∼≤ b−. Then a−∼+ b−∼ exists as well and we have a + b = a∼→ b−∼= a−∼+ b−∼. Since a∼ → b−∼ = (a∼ → b−∼)−∼, it follows that a + b ∈ Reg(A). Consequently,
ˆ
s(a + b) = s(a−∼+ b−∼) = s(a−∼) + s(b−∼) = ˆs(a) + ˆs(b).
Corollary 3.6.15 In any good pseudo-BCK-semilattice A, there is a one-to-one cor- respondence between the Rieˇcan states on A and those on Reg(A).
Let us return to Example 3.6.7. One readily sees that Reg(A) = {0, b, 1}. Actually, this is an MV-algebra with a unique state(-morphism), namely, m defined by m(0) = 0, m(b) = 12 and m(1) = 1 whose extension bm from Proposition 3.6.14 is just the Rieˇcan state s from Example 3.6.7. Consequently, the (pseudo-)BCK-semilattice A from our example admits a unique Rieˇcan state.
Chapter 4
Commutative
pseudo-BCK-algebras
Commutative pseudo-BCK-algebras are the “pseudo” version of well-known commu- tative BCK-algebras and (especially pseudo- LBCK-algebras) can be understood as a closer generalization of pseudo-MV-algebras in the setting of pseudo-BCK-algebras. We begin with basic properties of commutative pseudo-BCK-algebras and then concentrate mainly on commutative pseudo-BCK-algebra with the relative cancellation property (pseudo- LBCK-algebras for short). We show that they form a variety and that every pseudo- LBCK-algebra can be embedded into a pseudo-MV-algebra. Further, we inves- tigate some properties of pseudo- LBCK-algebras in comparison with the corresponding properties of the representing `-groups. In the final section we prove a Cantor-Bernstein type theorem for orthogonally σ-complete commutative pseudo-BCK-algebras.
4.1
Basic properties
A BCK-algebra (A, →, 1) is called commutative provided (x → y) → y = (y → x) → x for all x, y ∈ A. The analogue for pseudo-BCK-algebras offers immediately:
Definition 4.1.1 We say that a pseudo-BCK-algebra (A, →, , 1) is commutative if it satisfies the equations
(x → y) y = (y → x) x, (4.1.1)
(x y) → y = (y x) → x. (4.1.2)
It can be easily shown that in this case (A, ≤) is a join-semilattice, where
x ∨ y = (x → y) y = (x y) → y. (4.1.3)
Although it may seem to be a bit misleading (a commutative pseudo-BCK-algebra is not necessarily a BCK-algebra), we keep the term “commutative” as a natural genera- lization of commutative BCK-algebras. In [26] and [37], pseudo-BCK-algebras satisfy- ing (4.1.1) and (4.1.2) were considered under the name “semilattice-ordered” pseudo- BCK-algebras which, however, is confusing as well since a pseudo-BCK-algebra that is
a join-semilattice under its natural order need not fulfil the identities (4.1.1), (4.1.2) (see Example 2.1.9 or 2.1.14).
Let us recall two basic examples from Section 1.1; the examples are not prototypical, because there exist commutative pseudo-BCK-algebras that do not arise as subalgebras of (G−, →, , 1) or (M, →, , 1):
Example 4.1.2 Let (G, ·,−1, 1, ∧, ∨) be an `-group and let G− = {x ∈ G : x ≤ 1} be
its negative cone. If we equip G− with the operations → and defined by
x → y = y · (x ∨ y)−1,
x y = (x ∨ y)−1· y, (4.1.4)
then (G−, →, , 1) is a commutative pseudo-BCK-algebra.
Example 4.1.3 Let (M, ⊕,−,∼, 0, 1) be a pseudo-MV-algebra and define
x → y = x−⊕ y and x y = y ⊕ x∼. Then (M, →, , 1) is a commutative pseudo-BCK-algebra.
Lemma 4.1.4 The identities (4.1.1) and (4.1.2) are equivalent to the quasi-identities
x ≤ y ⇒ y = (y → x) x, (4.1.5)
x ≤ y ⇒ y = (y x) → x, (4.1.6)
respectively.
Proof: Let (A, →, , 1) be a pseudo-BCK-algebra satisfying (4.1.1). If x ≤ y then x → y = 1 and so y = 1 y = (x → y) y = (y → x) x.
Conversely, assume that (A, →, , 1) satisfies (4.1.5). Then x ≤ (x → y) y entails (x → y) y = (((x → y) y) → x) x. Further, y ≤ (x → y) y yields (y → x) x ≤ (((x → y) y) → x) x, and hence (y → x) x ≤ (x → y) y. By replacing x and y we obtain the other inequality.
The proof of the equivanlece of (4.1.2) and (4.1.6) is parallel.
It is evident that if (A, →, , 1) is a (commutative) pseudo-BCK-algebra, then for every a ∈ A, ([a, 1], →, , a, 1) is a bounded (commutative) pseudo-BCK-algebra. Since bounded commutative pseudo-BCK-algebras are termwise equivalent to pseudo- MV-algebras (see Example 1.1.4 and 1.1.8), we obtain
Proposition 4.1.5 Let (A, →, , 1) be a commutative pseudo-BCK-algebra. For any a ∈ A \ {1} define x ⊕ay = (x a) → y = (y → a) x, x−a = x → a and
x∼a = x a for x, y ∈ [a, 1]. Then ([a, 1], ⊕
a,−a,∼a, a, 1) is a pseudo-MV-algebra in
which x ·ay = (x → (y → a)) a = (y (x a)) → a.
There are no proper finite commutative pseudo-BCK-algebras:
Corollary 4.1.6 Let (A, →, , 1) be a commutative pseudo-BCK-algebra such that all sections [a, 1], a ∈ A, are finite. Then x → y = x y for all x, y ∈ A, thus (A, →, 1) is a commutative BCK-algebra.
4.1. Basic properties 67
Proof: For any x, y ∈ A we have (x ∨ y) → y = x → y and (x ∨ y) y = x y. In view of Proposition 4.1.5, ([y, 1], ⊕y,−y,∼y, y, 1) is a finite pseudo-MV-algebra. But
finite pseudo-MV-algebras are MV-algebras (see [18]), i.e., ⊕y is commutative and the unary operations−y and ∼y coincide, and consequently, likewise the restrictions of →
and to [y, 1] coincide. This entails x → y = (x ∨ y) → y = (x ∨ y) y = x y for all x, y ∈ A.
From Proposition 4.1.5 several corollaries concerning the natural order ≤ follow: Corollary 4.1.7 If (A, →, , 1) is a commutative pseudo-BCK-algebra, then for every a ∈ A \ {1}, ([a, 1], ≤) is a distributive lattice with
x ∨ay = x ∨ y,
x ∧ay = ((x → y) → (x → a)) a = ((x y) (x a)) → a.
Lemma 4.1.8 Let (A, →, , 1) be a commutative pseudo-BCK-algebra. For every x, y ∈ A, x ∧ y = infA{x, y} exists if and only if there is a ∈ A such that a ≤ x and
a ≤ y, and in this case, x ∧ y = x ∧ay.
Proof: Let a ≤ x and a ≤ y for some a ∈ A. We know that x ∧ay is a lower bound
of {x, y}. If z is another common lower bound of x and y then a ∨ z ≤ x and a ∨ z ≤ y. But a ∨ z ∈ [a, 1], and so z ≤ a ∨ z ≤ x ∧ay, showing x ∧ay = x ∧ y. Conversely, if
x ∧ y exists then we may obviously take a = x ∧ y.
Corollary 4.1.9 For (A, →, , 1) a commutative pseudo-BCK-algebra, if (A, ≤) is a lattice then it is a distributive lattice.
Proof: Let x, y, z ∈ A. By the previous lemma we can calculate x ∧ (y ∨ z) e.g. in the interval [a, 1] for a = x ∧ y ∧ z, which is a distributive lattice. Hence we have x ∧ (y ∨ z) = x ∧a(y ∨ z) = (x ∧ay) ∨ (x ∧az) = (x ∧ y) ∨ (x ∧ z).
This means that (A, ≤) is a (necessarily distributive) lattice if and only if every pair of elements of A has a common lower bound, i.e., if (A, ≤) is a directed poset.
In what follows we observe some properties of the class of all commutative pseudo- BCK-algebras. First of all, commutative pseudo-BCK-algebras form a variety since the quasi-identity (1.1.12) can be replaced by the identities (4.1.1) and (4.1.2). Indeed, if x → y = 1 and y → x = 1 then x = 1 x = (y → x) x = (x → y) y = 1 y = y.
We can give an alternative axiomatization which is similar to the axiomatization of commutative BCK-algebras (see [70] or [18]):
Proposition 4.1.10 An algebra (A, →, , 1) of type h2, 2, 0i is a commutative pseudo- BCK-algebra if and only if it satisfies the identities
(x → y) y = (y → x) x = (x y) → y = (y x) → x, (4.1.7)
x → (y z) = y (x → z), (4.1.8)
x → 1 = 1 = x 1, (4.1.9)
Proof: Let (A, →, , 1) be an algebra satisfying (4.1.7)—(4.1.10). It is easily seen that
x → x = 1 = x x
Indeed, by (4.1.10), (4.1.7) and (4.1.9), x → x = (1 x) → x = (x 1) → 1 = 1, and similarly, x x = 1. The identities (4.1.7) and (4.1.10) also capture the quasi-identity
x → y = 1 & y → x = 1 ⇒ x = y.
Indeed, if x → y = 1 and y → x = 1 then x = 1 x = (y → x) x = (x → y) y = 1 y = y.
We now prove that
x → y = 1 ⇔ x y = 1. (4.1.11)
If x → y = 1 then y = 1 y = (x → y) y = (x y) → y by (4.1.7), and hence 1 = y y = ((x y) → y) y = (y → (x y)) (x y) =
= (x (y → y)) (x y) = (x 1) (x y) = 1 (x y) = x y by (4.1.7) and (4.1.8). Thus x → y = 1 implies x y = 1, and analogously, x y = 1 entails x → y = 1.
It remains to verify the identities (1.1.7). We have:
(x → y) → [(y → z) (x → z)] = (x → y) → [x → ((y → z) z)]
= (x → y) → [x → ((z → y) y)] = (x → y) → [(z → y) (x → y)] = (z → y) [(x → y) → (x → y)] = (z → y) 1 = 1,
so (x → y) [(y → z) (x → z)] = 1 by (4.1.11); the identity (1.1.8) is obtained analogously.
Theorem 4.1.11 The variety C of commutative pseudo-BCK-algebras is weakly regu- lar, congruence distributive and 3-permutable. In addition, the variety Lc of all com-
mutative pseudo-BCK-lattices is arithmetical and regular.
Proof: By Theorem 2.2.10, the variety C is weakly regular and distributive. In order to show 3-permutability we have to find ternary terms t1 and t2 with t1(x, y, y) = x, t1(x, x, y) = t2(x, y, y) and t2(x, x, y) = y (see e.g. [6]). It suffices to take t1(x, y, z) = (z → y) x and t2(x, y, z) = (x → y) z.
The variety Lc is a subvariety of the variety L of all pseudo-BCK-lattices which
is arithmetical by Theorem 3.1.3. For the final claim, we must find terms p1, . . . , pn (n ∈ N) such that
p1(x, y, z) = · · · = pn(x, y, z) = z iff x = y.
Let n = 2 and
p1(x, y, z) = (x → y) ∧ (y → x) ∧ z,
4.2. Relative cancellation property 69