Decidability of the equational theory of meadows

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Decidability of the equational theory of meadows

Sebastian Zur

August 26, 2016

Supervisor: mw. dr. I. Bethke, dhr. dr. G. Regts

Korteweg-de Vries Instituut voor Wiskunde

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Abstract

In this thesis we will construct an algorithm to show that the equational theory of meadows is decidable. We will begin by looking at the axioms of commutative von Neumann regular rings and expanding these rings with a multiplicative inverse. The inverse in commutative rings is usually axiomatized by:

∀x∃x−1((x 6= 0) → xx−1= 1).

We will however use an alternative axiom system that does not use quantifiers as introduced in [1]. The algebraic structures that satisfy this new set of axioms are called meadows [2].

To prove that the equational theory of meadows is decidable means to prove that the validity of every equation in a meadow can be checked in a finite number of steps. In Chapter 2 we will use model theory to show that it is enough to look at the just the meadows that also satisfy the axioms of algebraically closed fields. Our algorithm consists of two parts: rewriting the equation in an universal formula and removing the quantifiers in this universal formula. We will first discuss these part in a theoretical way in Chapters 3 and 4. After having mathematically supported these steps of our algorithm, we shall give the actual algorithm in a constructive way in Chapter 5, together with its complexity.

Title: Decidability of the equational theory of meadows

Authors: Sebastian Zur, zursebastian@student.uva.nl, 10547800 Supervisor: mw. dr. I. Bethke, dhr. dr. G. Regts

Second grader: dhr. dr. A. Ponse, dhr. dr. L. Taelman Date: August 26, 2016

Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam

Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math

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1 Introduction

1.1 Variety

Fundamental mathematic theorems like the Pythagoras theorem, Fermat’s little theorem and L’Hˆopital’s rule have been proved hundreds of years ago. In those times the mathematicians only had their own mind to their disposal.

With the emergence of computers however, a subfield of mathematical logic has been created called Automated Theorem Proving (ATP). This subfield deals with algorithms that show where a certain statement is true using only a finite set of axioms. For ATP it is more amenable if these axioms are equational axioms (axioms of the form a = b, since this allows the algorithm the use of simple substitutions). This thesis will use a lot of terminology from model theory [3]:

Definition 1.1.1. An algebraic structure is a set with one or more operators defined on it that satisfy a list of axioms.

These operators are functions from the set to itself that take a finite number of elements and map onto one element. The number of elements this operator takes as its input is called the arity of the operator. A n-ary operator takes n elements as its input. The arity of an operator can also be zero, then the operator is a constant. We will now explain some model theoretic terminology:

The set of operators that work on a algebraic structure will be denoted as the language of the structure. Now let ∆ be a set of axioms. If an algebraic structure satisfies all these axioms, this algebraic structure is a model of ∆.

As said before, we are only interested in equational axioms for ATP, so if we want to use a certain algebraic structure in ATP, this structure will have to be axiomatized by equational axioms.

Definition 1.1.2. A variety (not to be confused with a variety in analysis or algebraic geometry) is the class of all algebraic structures with the same language that are a model of a finite set of equational axioms.

Example 1.1.3. An example of a variety is the set of all commutative groups. Normally this algebraic structure is defined with a single operation + and the following axioms:

1. ∀x, y, z x + (y + z) = (x + y) + z. 2. ∀x, y x + y = y + x.

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4. ∀x ∃i such that x + i = e.

Even though the second and third axiom are not equational axioms, they can be rewritten by introducing a nullary operator e and a unary operator − and defining them as follows:

1. ∀x, y, z x + (y + z) = (x + y) + z. 2. ∀x, y x + y = y + x.

3. ∀x e + x = x. 4. ∀x x + (−)x = e.

Where the set of all groups is a variety, the set of all fields is not. The multiplicative inverse is not defined for the additive identity element and thus it can not be rewritten into an equational axiom that is true for every element (including the additive identity element). Therefore Bergstra [4] has introduced meadows as a purely equational axiomatized alternative for fields.

1.2 Meadows

The axiom system Σ for commutative von Neumann regular rings (from now on we will use the word ‘regular’ to abbreviate von Neumann regular) consists of two parts, the first part being the axioms for commutative rings:

1. ∀x, y, z x + (y + z) = (x + y) + z. 2. ∀x, y x + y = y + x. 3. ∀x 0 + x = x. 4. ∀x x + (−)x = 0. 5. ∀x, y, z x ∗ (y ∗ z) = (x ∗ y) ∗ z. 6. ∀x, y x ∗ y = y ∗ x. 7. ∀x 1 ∗ x = x. 8. ∀x, y, z (x + y) ∗ z = x ∗ z + y ∗ z. 9. ∀x, y, z x ∗ (y + z) = x ∗ y + x ∗ z.

The + and ∗ are binary operators, the − is an unary operator and the 1 and 0 are nullary operators.

The second part consist of the following axiom:

∀x ∃a, such that x = x ∗ a ∗ x. (1.1) Note that x 6= 0 → a 6= 0 if we also work in an integral domain.

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Remark 1.2.0.1. From now on we will write a + (−)b as a − b.

Lemma 1.2.1. Any integral domain F which is a model of Σ is a field.

Proof. A integral domain is a commutative ring where the product of two nonzero elements is nonzero. So if an integral domain is a model of Σ, we only add the axiom of regularity (1.1) to the axiom set of an integral domain. This axiom can be rewritten to:

∀x ∃a, such that x ∗ (1 − a ∗ x) = 0. (1.2) Since we are working in an integral domain, either x = 0 or a ∗ x = 1. So every nonzero element has a multiplicative inverse and thus F is not only a integral domain but more specifically a field.

Let L be the language for rings, which consists of our previous defined operators {+, −, ∗, 0, 1}. This language can be extended by the unary operator −1 to form the language L0. This language will be used for models of Σ. This operator is defined as follows:

∀x x ∗ x−1∗ x = x.

∀x (x−1)−1= x. (1.3) Lemma 1.2.2. For commutative rings (1.1) and (1.3) are equivalent

Proof. This proof will consist of two parts.

First assume that (1.1) holds. This implies that for every element x in a commutative ring there exists an a such that x = x ∗ a ∗ x. Now assume that a 6= a ∗ x ∗ a and x 6= 0:

a 6= a ∗ x ∗ a. x ∗ a ∗ x 6= x ∗ a ∗ x ∗ a ∗ x.

x 6= x ∗ a ∗ x.

This gives a contradiction and thus a 6= a ∗ x ∗ a if x 6= 0. For the case that x = 0 there exists a = 0 such that x = x ∗ a ∗ x and a = a ∗ x ∗ a. This a is unique for every x. Since if there exists another b such that x = x ∗ b ∗ x holds, we have that b = a:

b = b ∗ x ∗ b = b ∗ x ∗ a ∗ x ∗ b = b ∗ a ∗ x. a = a ∗ x ∗ a = a ∗ x ∗ b ∗ x ∗ a = a ∗ b ∗ x = b. a = b.

Thus for every x there exists an unique x−1 = a that satisfies (1.3).

The second part is easier. Assume that (1.3) holds, then choose x−1 as a in (1.1).

Thus by changing (1.1) into (1.3) in Σ, the axiom set is now purely equational and the class of all algebra¨ıc structures that satisfy Σ is a variety. These algebra¨ıc structures are called meadows.

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Corollary 1.2.2.1. For any meadow, the following is true: • 0−1 = 0.

• x ∗ x−1 = 0 ↔ x = 0. • x ∗ y = 1 → y = x−1_.

• (x ∗ y)−1 = x−1∗ y−1.

Proof. In the following proof the second axiom in (1.3) is checked by checking if x−1= x−1∗ x ∗ x−1 holds. Note that this holds for all commutative rings.

• Since multiplying with zero always results in zero, we have that 0−1_{= 0}−1_∗0∗0−1 ₌

0 and 0 = 0 ∗ 0−1 = 0. So 0−1= 0.

• If x ∗ x−1 _{= 0, then x = x ∗ x}−1_{∗ x = 0. The converse is trivial.}

• If x ∗ y = 1, then x = x ∗ y ∗ x and y = y ∗ x ∗ y. So y = x−1. • Since meadows are commutative we have that:

(x ∗ y) ∗ (x ∗ y) ∗ (x−1∗ y−1) = x ∗ x ∗ x−1∗ y ∗ y ∗ y−1 = x ∗ y and

(x ∗ y) ∗ (x−1∗ y−1) ∗ (x−1∗ y−1) = x ∗ x−1∗ x−1∗ y ∗ y−1∗ y−1 = x−1∗ y−1 We conclude that (x ∗ y)−1 = x−1∗ y−1_.

Meadows seem a suitable alternative of fields for the use in ATP, but the class of all meadows being a variety is not enough. The equational theory of the meadows also has to be decidable. This means that there should exist an algorithm that can decide whether an equation in a meadow is true or false in a finite number of steps. These steps can consist of arithmetic operations or substitutions using the axioms. From now on with a/b we mean a ∗ b−1 and ab means a ∗ b.

1.3 Examples of meadows

Even though in the rest of this thesis we will only discuss the class of all meadows instead of specific examples of meadows, these examples will clarify the concept of meadows. Example 1.3.1. It follows from Σ that expanding any field with 0−1 = 0 results in a meadow.

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Example 1.3.2. Consider the ring Z/6Z. This ring is a meadow since: 0−1 = 0 3−1 = 3

1−1 = 1 4−1 = 4 2−1 = 2 5−1 = 5

There is exists a more general theorem for whether a ring of integers modulo n is a meadow:

Theorem 1.3.3. Z/nZ is a meadow if and only if n is the product of distinct primes (squarefree)

Proof. We shall follow the proof in [5]. In this proof we always assume that x, a 6≡ 0 mod n, as every commutative ring has 0−1∗ 0. Assume that Z/nZ is a meadow. This means that for every x there is a unique a such that:

x ≡ x2a mod n a ≡ a2x mod n.

Now let pk divide n for some k ≥ 1. Substituting x = p in the first equation results in k = 1, since if k ≥ 2, then p2a mod n = 0 6= p.

For the converse assume that n is squarefree. Denote gcd(x, n) as the greatest common divisor of x and n. If gcd(x, n) = 1, then x has an inverse in Z/nZ and Z/nZ is a meadow.

Now assume that gcd(x, n) > 1. The assumption is that n is squarefree, so n = p1...pm

with p1, ..., pm different primes. This implies that gcd(x, n) = Qi∈Mpi = pM, where

M ⊂ {1, ..., m}. We can write x = ypM, where pM does not divide y.The first equation

can be rewritten to ypM ≡ y2p2Ma mod n, i.e:

y = y2pMa mod (pMc).

Since gcd(y2pM, pMc) = 1, y2p_M has an inverse in Z/nZ and we can divide by y2p_M to

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2 Sets of equations

2.1 Birkhoff ’s theorem

Before we show that the equational theory of meadows is decidable, we will show that it suffices to look at algebra¨ıcally closed meadows. One of the most fundamental theorems in model theory is Birkhoff’s theorem [6]:

Theorem 2.1.1 (Birkhoff’s theorem). A class of algebraic structures of the same language is a variety if and only if that class is closed under the taking of homomorphic images, subalgebras and direct products.

In this context an homomorphic image is a function from a set to another set that preserves the operations of the respective set.

The proof of this theorem is quite complex, so we will only proof a small part of this theorem, but first we will discuss some more model theoretic terminology:

Definition 2.1.2. A term t of a language A is a finite combinations of variables and operators in A.

For the language L this corresponds with polynomials, here are some examples of terms in L: t1 = x1+ 25x0, t2= 5, t3= 12 X i xi.

Definition 2.1.3. An equation of a language A is a formula of the form t = s, where t and s are terms of A. An equation holds in a model B, if it holds for all assignments of values in B to the variables in the equation. We will denote the set of all equations which hold in a ring B as Eq(B).

Now we can proof a small lemma that is a part of 2.1.1:

Lemma 2.1.4. Let V be a model of some purely equational axiom set ∆ and g an homomorphic image. Then g(V ) is axiomatized by ∆.

Proof. We will follow the proof in [6]. Since every axiom in ∆ is an equation, every axiom in ∆ can be written in the following form

p(x1, ..., xn) = q(x1, ..., xn).

Now take b1, ..., bn ∈ g(V ), then there exist a1, ..., an ∈ V such that g(ai) = bi, with

1 ≤ i ≤ n

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As we can apply g on both sides, this results in

g(p(a1, ..., an)) = g(q(a1, ..., an)).

Since operations are preserved under taking homomorphic images, this equality implies p(g(a1), ..., g(an)) = q(g(a1), ..., g(an)).

p(b1, ..., bn) = q(b1, ..., bn).

As this holds for all b1, ..., bn ∈ g(V ) and for all axioms in ∆ we conclude that g(V ) is

also a model of ∆.

To minimize notation, we will follow the terminology defined in [1]:

Let Γ be a set of equations, then mod Γ is the set of all models of Γ. So mod Γ contains all the algebraic structures that satisfy Γ. For instance mod Σ equals the class of all meadows.

A subset of mod Γ is mod∗ Γ. This set consists of all models of Γ which also satisfy the axioms of fields.

We will now show that equations that hold in all models of Σ also hold in all fields that satisfy Σ and vice versa.

2.2 Equivalence of sets of equations

Definition 2.2.1. A subdirect product is a subalgebra S of a direct productQ

i∈ISi,

if there exists a injective homomorphism h : S →Q

iSi such that πi◦ h is a surjective

homomorphism for each i, with πi being the canonical map fromQ_i∈ISi to Si.

Theorem 2.2.2. Let Γ be a set of equations with Σ ⊆ Γ. Then every commutative regular ring R in mod Γ is the subdirect product of fields {Si}i∈I in mod∗ Γ.

Proof. The idea of this proof is based on proofs in [7] and [8]. Let R be a commutative regular ring R in mod Γ. Then R satisfies (1.2) and thus for all a 6= 0 in R we can find a x such that 1 − xa is a zero-divisor. Let Si = R/Ki, with {Ki}i∈I all the prime ideals of

R. Since R is commutative, {Ki}i∈I are also all the maximal ideals of R.

Suppose that there exists an a 6= 0 in T

i∈IKi. Then every maximal ideal Ki contains

a. For every x in R, Ki should contain xa, as Ki is an ideal. Assume that 1 − xa is an

element of Ki. This would imply that 1 is an element of Ki. Since 1 is not an element

of any maximal ideal, we have a contradiction and thus 1 − xa is not an element of any Ki.

As such, the ideal I generated by 1 − xa is not a maximal ideal. This means that either I is included in a maximal ideal or that I contains 1. As we have already shown, 1 − xa is not in any maximal ideal and thus the second case must be true. This implies however that (1 − xa)s = 1 for some s ∈ R and thus 1 − xa is a unit for every x in R. This gives a contradiction with the statement that 1 − xa is a zero-divisor for a certain x

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in R and thus T i∈IKi = 0. We now define h : R →Q iSi = Q iR/Ki with h(r) = Q

ir mod Ki. It is clear that h

is an homomorphism. If we look at the kernel of this image, we see that h(r) = 0 implies that r mod Ki= 0 for all Ki. This implies that r is an element of all Ki and as shown

above this means that r = 0. We conclude that h is a injective homomorphism.

Combining h with the canonical map πi results in πi◦ h, the natural quotient map from

R to R/Ki and therefore a surjective homomorphism. So R is the subdirect product

of Si. As Ki are maximal ideals, these Si are fields. Because of 2.1.4 these fields are a

model of Γ, since they are the image of R under homomorphic images. Corollary 2.2.2.1. Continue the notation in 2.2.2, then Eq(R) =T

i∈I Eq(Si).

Proof. Since every Si is the image of R under the surjective homomorphism πi ◦ h,

theorem gives us that Eq(R) ⊆ Eq(Si) for all i in I and thus Eq(R) ⊆ Ti∈I Eq(Si).

Now assume that an equation holds forT

i∈I Eq(Si). This is equivalent to saying that

the equation holds for all Si. It follows that the equation holds for the direct product

Q

i∈ISi:

Corollary 2.2.2.2. Let Γ be a set of equations, with Σ ⊆ Γ. Then Eq(mod Γ) = Eq(mod∗ Γ) holds.

Proof. It is clear that Eq(mod Γ) ⊆ Eq(mod∗ Γ), since if an equation holds for all models of Γ, it also holds for a subset of these models. Say an equation e holds for all meadows in mod Γ. For all models R of Γ, we know according to 2.2.2, that Eq(R) =T

i Eq(Si),

with Si meadows in mod∗ Γ. So e holds for all models R of Γ and so Eq(mod∗ Γ) ⊆

Eq(mod Γ).

Let us introduce mod∗∗Γ as the set of all models of Γ which are algebraically closed fields. Recall that a field F is algebraically closed if all roots of any polynomial in F [x] are elements of F . We will now give a stronger version of 2.2.2.2:

Corollary 2.2.2.3. Let Γ be a set of equations, with Σ ⊆ Γ. Then Eq(mod Γ) = Eq(mod∗∗ Γ) holds.

Proof. It is clear that mod∗∗ Γ is a subset of mod∗ Γ, and thus Eq(mod∗ Γ) ⊆ Eq(mod∗∗ Γ).

Every field is a subalgebra of its algebraic closure. Since every field has an algebraic closure, we conclude using 2.1.1 that Eq(mod∗∗ Γ) ⊆ Eq(mod∗ Γ).

Combining this with 2.2.2.2 results in Eq(mod Γ) = Eq(mod∗∗ Γ).

This result is essential for the algorithm described in Chapter 4. It is no longer needed to check whether an equation holds in all meadows, but only in all algebraic closed fields (these fields still have to be models of Σ). As such from now on we assume that we work

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3 Universal formula

3.1 Standard Meadow Form

Instead of giving an existential proof of the decidability of the equational theory of meadows, we will give an algorithm that can check whether an equation does or does not hold in all meadows. The first step of this algorithm consists of showing that every equation can be rewritten into a universal formula (this will be defined later). To proof that this is possible, we will use a standard representation for meadows introduced by in [2].

Definition 3.1.1. Let t be a term of the language L0. This term is a Standard Meadow Form (SMF) if t is a SMF of level n for some n ∈ N . These levels are defined as:

SMF of level 0: each expression of the form t1/t2, with t1, t2 terms of the language L.

SMF of level n: each expression of the form (t1/t1)P + (1 − t1/t1)Q, with t1

a term of the language L and P, Q expressions with SMF level n − 1. Analogous to polynomials, if a term t is a SMF of level n, then it is also a SMF of level n + k, where k is a natural number.

Even though we can assume that we work on algebraically closed fields and thus t/t = 1 if t 6= 0 and t/t = 0 otherwise, we will still prove this lemma for all meadows. As such, we only know the following:

(t/t)t = t. (t/t)(1/t) = (1/t). (t/t)(t/t) = t/t. (t/t)(1 − t/t) = 0. (1 − t/t)t = 0. (1 − t/t)(1/t) = 0. (1 − t/t)(1 − t/t) = 0.

We will not prove these equalities, as they are easily derived from Σ. Lemma 3.1.2. If P, Q are SMFs then:

1. P + Q is a SMF with the same variables. 2. P Q is a SMF with the same variables.

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3. −P is a SMF with the same variables. 4. 1/P is a SMF with the same variables.

Proof. If P, Q are SMFs of level 0, then P = t1/t2 and Q = t3/t4, with t1, ..., t4 terms of

the language L. 1.

P + Q = (t2/t2)(P + Q) + (1 − t2/t2)(P + Q)

= (t2/t2)(t1/t2+ t3/t4) + (1 − t2/t2)(t1/t2+ t3/t4)

= (t2/t2)(t1/t2+ (t2/t2)(t3/t4)) + (1 − t2/t2)(t3/t4).

What rests to prove is that t1/t2+ (t2/t2)(t3/t4) is a SMF:

t1/t2+ (t2/t2)(t3/t4) = (t4/t4)(t1/t2+ (t2/t2)(t3/t4)) + (1 − t4/t4)(t1/t2+ (t2/t2)(t3/t4))

= (t4/t4)((t4/t4)(t1/t2) + (t2/t2)(t3/t4)) + (1 − t4/t4)(t1/t2)

= (1 − t4/t4)(t1/t2) + (t4/t4)((t1t4+ t2t3)/(t2t4)).

So t1/t2+ (t2/t2)(t3/t4) is a SMF of level 1 and thus P + Q is a SMF of level 2.

2. P Q = (t1/t2)(t3/t4) = (t1t3)/(t2t4), so P Q is a SMF of level 0.

3. −P = −t1/t2 = (−t1)/t2, so −P is a SMF of level 0.

4. 1/P = 1/(t1/t2) = t2/t1, so 1/P is a SMF of level 0.

Suppose that 3.1.2 holds for SMFs of level n.

Now let P, Q be SMFs of level n + 1, then P = (t1/t1)S + (1 − t1/t1)T and Q =

(t2/t2)U + (1 − t2/t2)V , with S, T, U, V SMFs of level n: 1. P + Q = (t1/t1)P + (1 − t1/t1)P + Q = (t1/t1)(S + Q) + (1 − t1/t1)(T + Q) = (t1/t1)((t2/t2)(S + U ) + (1 − t2/t2)(S + V ))+ (1 − t1/t1)((t2/t2)(T + U ) + (1 − t2/t2)(T + V )).

And this is a SMF because of the induction hypothesis. 2.

P Q = (t2/t2)P U + (1 − t2/t2)P V

= (t2/t2)((t1/t1)SU + (1 − t1/t1)T U )+

(1 − t1/t1)((t2/t2)SV + (1 − t2/t2)T V ).

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3.

−P = −(t1/t1)S + (1 − t1/t1)T )

= t1/t1(−S) + (1 − t1/t1)(−T ).

And this is a SMF because of the induction hypothesis. 4. 1/P = (t1/t1)(1/P ) + (1 − t1/t1)(1/P ) = t1/t1( 1 (t1/t1)S + (1 − t1/t1)T ) + (1 − t1/t1)( 1 (t1/t1)S + (1 − t1/t1)T ) = t1/t1( t1/t1 (t1/t1)(t1/t1)S + (t1/t1)(1 − t1/t1)T ) + (1 − t1/t1)( 1 − t1/t1 (1 − t1/t1)(t1/t1)S + (1 − t1/t1)(1 − t1/t1)T ) = t1/t1(1/S) + (1 − t1/t1)(1/T ).

And this is a SMF because of the induction hypothesis.

With these properties we can see that every term t of the language L0 can be written as a SMF and thus we can assume that a term is written as a SMF.

Remark 3.1.2.1. If we work on a field, t/t = 1 if and only if t 6= 0. As such, all terms in a field can be written as a SMF of level 0, since (t1/t1)P + (1 − t1/t1)Q is either P or Q.

3.2 Universal formulas

In predicate logic, we look at formulas. These are a finite combination of variables, logical connectives and quantifiers. Since these formulas have a truth value (either true or false), they always contain the = symbol in model theory. Apart from this symbol they may also contain variables and operators from the language in which they are set. There are various forms of formulas:

Definition 3.2.1. An atomic formula is a formula that contains no logical connectives nor quantifiers.

Example 3.2.2. Since variables and operators are still allowed in atomic formulas, an example of an atomic formula in the language L would be x1+ 1 = 0.

Definition 3.2.3. An open formula is a formula that contains no quantifiers.

Example 3.2.4. Logical connectives are allowed in open formulas, so we can add logical connectives to 3.2.2: (x1= 0) ∨ (x2 = 0).

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Definition 3.2.5. An universal formula is a formula that does not contain the existential quantifier.

Example 3.2.6. As the universal quantifier is allowed in an universal formula, we can add it to 3.2.4: ∀x1, x2((x1 = 0) ∨ (x2 = 0)).

Theorem 3.2.7. Let t be a term of the language L0 with variables x1, ..., xn and let y

be a variable not appearing in t. Then there exists an effective method of getting an open formula ψ(x1, ..., xn, y) with variables x1, ..., xn, y such that:

y = t ↔ ψ(x1, ..., xn, y).

where every atomic formula in ψ(x1, ..., xn, y) is either of the form s = 0 or y = s0, where

s is a term in L and s0 is a SMF of level 0.

Proof. If t is a SMF of level 0 then ψ(x1, ..., xn, y) = (y = t) is an open formula of the

desired form.

Now we assume the theorem holds for SMFs of level n. Let t be a SMF of level n + 1, then t = (s/s)P + (1 − s/s)Q, where s is a term of L and P, Q are SMFs of level n.

y = t ↔ y = s/s ∗ P + (1 − s/s) ∗ Q

↔ (s = 0 ∧ y = Q) ∨ (¬(s = 0) ∧ y = P ).

Assuming that the theorem holds for SMFs of level n, we know that there exist open formulas ψP(x1, ..., xn, y) and ψQ(x1, ..., xn, y) of the desired form with ψP(x1, ..., xn, y) ↔

y = P and ψQ(x1, ..., xn, y) ↔ y = Q. So

ψ(x1, ..., xn, y) = (s = 0 ∧ ψQ(x1, ..., xn, y)) ∨ (¬(s = 0) ∧ ψP(x1, ..., xn, y))

is an open formula of the desired form.

We have now proven that any term of the language L0 can be written as an open formula. This gives a way of getting a open formula ψt1=t2(x1, ..., xn): We take the

formula received from 3.2.7 for the equation y = t1 − t2 and replace all the formulas

where y is a SMF of level 0 (y = s1/s2) with s1 = 0 ∨ s2= 0. We shall call t1− t2= 0

our original equation. Note that this results in an open formula of language L!

Since we want our original equation to be true for every assignment of variables we have to verify whether (∀x1, ..., xn)ψt1=t2(x1, ..., xn) is true. Note that this is not an open

formula anymore but an universal formula, since it contains the universal quantifier. An universal formula has a truth value just like equations. We shall denote T∀(R) as

the set of all universal formulas that hold in R, where R is an algebraically closed field in Σ. We have just shown that we can check whether any equations is an element of Eq(mod∗∗ Γ) ( = Eq(mod Γ)) by checking whether the corresponding universal formula is an element of T∀(mod∗∗ Γ).

Theorem 3.2.8. For any R in mod∗∗ Γ, Eq(R) is decidable if and only if T∀(R) is

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Proof. We have just shown that Eq(R) is decidable if T∀(R) is decidable. As the converse

will not be used in this thesis, the proof is not given. It can be found in [1].

Verifying whether an universal formula ∀x1, ..., xn(ψt1=t2(x1, ..., xn)) holds in R is a

problem, since R is an algebraically closed and thus by definition not finite.

As decidability does require a finite algorithm, the universal formula will have to be rewritten into an equivalent formula without quantifiers. This is done in the next chapter using quantifier elimination.

This chapter only provided an existential prove that every equation can be rewritten to an universal formula. In Chapter 5 we will show a constructive method to create this universal formula, such that it can be used in our algorithm.

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4 Quantifier elimination

4.1 Prenex normal form

We will show the method to remove quantifiers in universal formulas of algebraically closed fields as explained in [9]. It begins by putting the formula in its prenex normal form. ψ is in prenex normal form if it is of the following form:

(Q1y1)...(Qmym)φ(y1, ..., ym, x1, ..., xn). (4.1)

where y1, ..., ym, x1, ..., xn are variables, Q1, ..., Qm are quantifiers and φ is an open

for-mula.

Before we continue recall de-Morgan’s laws (where ψ1, ψ2 are universal formulas):

(ψ1∨ ψ2) ↔ ¬(¬(ψ1) ∧ ¬(ψ2)).

(ψ1∧ ψ2) ↔ ¬(¬(ψ1) ∨ ¬(ψ2)).

Lemma 4.1.1. Every universal formula ψ(x1, ..., xn) is logically equivalent to an universal

formula ˆψ(x1, ..., xn) in prenex normal form.

Proof. If ψ is an atomic formula, then ψ is already in prenex normal form and thus of form has form (4.1).

1. ¬ψ = (Q0₁y1)...(Q0nym)¬φ(y1, ..., ym, x1, ..., xn), with Q0i = ∃ if Q0i = ∀ and vice

versa.

2. Assume ψ0also has form (4.1) with quantifiers Q0₁, ..., Q0_nand variables y₁0, ..., y0_m, x0₁, ..., x0_n and φ0 instead of φ. Then

ψ∨ψ0 = (Q1y1)...(Qnym)(Q01y 0 1)...(Q 0 ny 0 m)(φ(y1, ..., ym, x1, ..., xn)∨φ0(y01, ..., y 0 m, x 0 1, ..., x 0 n)).

It is trivial that if φ is in prenex normal form, that then both (∃Yn+1)φ and (∀Yn+1)φ are

in prenex normal form as well. Combine this with the fact that the logical connectives ¬, ∨ are sufficient to create every universal formula and we have proven that every universal formula can be written in prenix normal form.

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4.2 Algorithm

The algorithm for removing quantifiers in universal formulas of an algebraically closed field R will inductively remove one quantifier at at time. We will denote the corresponding variable of the quantifier that we remove as y and the rest of the variables as ˆx.

If our universal formula looks like (4.1) and we want to remove quantifier Q3, then y = y3

and ˆx = y1, y2, y4, ..., ym, x1, ..., xn.

Before we give the actual algorithm, let us first prove the following lemma:

Lemma 4.2.1. Let p(ˆx, y) = ak(ˆx)yk+ ... + a0(ˆx) and pj(ˆx, y) = aj(ˆx)yj+ ... + a0(ˆx) be

polynomials in R[ˆx, y], where 0 ≤ j ≤ k and aj(ˆx) 6= 0. Every polynomial q(ˆx, y) can be

divided by pj(ˆx, y) with degy(pj(ˆx, y)) = j ≤ degy(q(ˆx, y)) = d, which produces qj(ˆx, y)

and rj(ˆx, y) in the following way:

aj(ˆx)dq(ˆx, y) = qj(ˆx, y)pj(ˆx, y) + rj(ˆx, y), (4.2)

where degy(rj(ˆx, y)) < degy(pj(ˆx, y)).

Proof. We will proof this by induction on d − j, where we will use the following notation: q = q(ˆx, y) = bd(ˆx)yd+ ... + b0(ˆx) = bdyd+ ... + b0. p = p(ˆx, y) = ak(ˆx)yk+ ... + a0(ˆx) = akyk+ ... + a0. qj = qj(ˆx, y). pj = pj(ˆx, y). rj = rj(ˆx, y). Let d − j = 0.

Choosing qj = ad−1_j bd and rj = ad−1_j (aj(bd−1yd−1+ ... + b0) − bd(aj−1yj−1+ a0)) results

in:

qjpj+ rj = (ad−1j bd)(ajyj+ ... + a0) + ad−1j (aj(bd−1yd−1+ ... + b0) − bd(aj−1yj−1+ a0))

= ad_jbdyd+ ajd−1bd(aj−1yj−1+ ... + a0) + adj(bd−1yd−1+ ... + b0) − bd(aj−1yj−1+ ... + a0)

= ad_j(bdyd+ ... + b0)

= ad_jq.

where d − 1 = degy(rj) < degy(pj) = d.

Let us assume that for d − j = n < d we can find qj and rj such that 4.2.1 holds.

Now let d − j = n + 1 ≤ d.

ad_jq = ad_j(bdyd+ ... + b0) = adj(bd−1yd−1+ ... + b0) + ad−1_j bdyd−j(ajyj).

This can be rewritten to:

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Define ˜q as:

aj(bd−1yd−1+ ... + b0) − bdyd−1(aj−1yj−1+ ... + a0).

Then ˜q is a polynomial of degree d − 1 in y. This implies that d − j = n and because of our assumption there exist ˜qj and ˜rj such that ad−1j q = ˜˜ qjpj+ ˜rj, with degy(˜r) <

degy(pj).

Choosing qj = ˜qj + ad−1j bdyd−1 and rj = ˜rj results in:

qjpj + rj = (˜qj + ad−1_j bdyd−1)pj+ ˜rj

= ˜qjpj+ ad−1j bdyd−1pj+ ˜rj

= ad−1_j q + a˜ d−1_j bdyd−1pj

= ad_jq, where degy(r) = degy(˜r) < degy(pj).

Since ∀(ψ1) ↔ ¬(∃(¬(ψ1))), it is enough to remove the ∃ quantifier. Continue the use

of the notation in (4.1), then removing a quantifier from ψ is the same as removing a quantifier from the open formula φ. Every open formula can be written as the disjunction, conjunction and/or negation of open formulas. Combining this with de-Morgan’s law and the fact that (∃y)(φ1∨ φ2) ≡ (∃y)(φ1) ∨ (∃y)(φ2), we are reduced to only having to

consider formulas of the following form: (∃y)(^

i∈I

ti(ˆx, y)), (4.3)

where ti(ˆx, y) is either an atomic formula or the negation of one.

At the end of Chapter 3 we concluded that the universal formula equivalent to our original equation was set in the language L. As such, every atomic formula ti is of the

language L. As these atomic formulas do not contain any logical connectives and/or quantifiers, but only variables and operator in L, they can be viewed as polynomials in R[ˆx, y], where R is the algebraically closed field we work on.

So every ti is equivalent to a polynomial either equal or unequal to zero. Also, since we

work on a field, x 6= 0 ∧ y 6= 0 is equivalent to xy 6= 0. As such, (4.3) is equivalent to: (∃y)(f1= 0 ∧ ... ∧ fm= 0 ∧ g 6= 0), (4.4)

with f1, ..., fm, g polynomials in R[ˆx, y].

The actual removal of the quantifier ∃ from formulas of the form (4.4) consists of three parts:

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4.2.1 Part 1

In this part we will reduce (4.4) to the case where only one of the fi contains y. We

can assume that degy(fi(ˆx, y)) ≥ 1 for i = 1, ..., m and m ≥ 1. We have namely that

(∃y)(φ1∧ φ2) is equivalent to φ1∧ (∃y)(φ2) if ψ1does not contain y. We will use induction

on Pm

i=1 degy(fi(ˆx, y)):

We continue the use of notation in 4.2.1. Choose two of our fi as p and q.

Since p = 0 ∧ q = 0 is equivalent to pj = 0 ∧ rj = 0 if

ak= ... = aj+1 = 0 and aj 6= 0, we can rewrite p = 0 ∧ q = 0 as follows: k _ j=0 (ak= 0 ∧ ... ∧ aj+1= 0 ∧ aj 6= 0 ∧ pj = 0 ∧ rj = 0) ∨ (a_k= 0 ∧ ... ∧ a0 = 0 ∧ q = 0). (4.5)

Again use (∃y)(φ1∨ φ2) ↔ (∃y)(φ1) ∨ (∃y)(φ2) to split this formula at the conjunction

connectives. Since degy(pj) ≤ degy(p) and degy(rj) ≤degy(q), we reduce Pm_i=1 degy(fi).

Since the ai = 0 atomic formulas do not contain the variable y, after repeatedly applying

part 1 the only formulas that contain the quantifier y are of the form:

(∃y)(u = 0 ∧ v 6= 0), (4.6) with u, v polynomials in R[ˆx, y].

As repeatedly applying part 1 decreases Pm

i=1degy(fi) with every step, eventually we do

not have two polynomials equal to zero anymore and thus we cannot apply the above described method anymore.

Remark 4.2.1.1. When looking correctly at (4.5), the formulas might also have the form:

(∃y)(v 6= 0). (4.7) These will be discussed in Part 3 of the algorithm.

4.2.2 Part 2

We have now arrived at the point that we have to remove the y in (4.6).

Let ˜x be any assignment of values for ˆx in R. Since we work on an algebraically closed field, we have the following:

u(˜x, y) = (y − α1)...(y − αk).

v(˜x, y)k= (y − β1)k...(y − βl)k.

with αi, βj elements of R. The statement

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is equivalent to there existing an αi such that αi is unequal to all βj. This ofcourse

implies that u(˜x, y) not does not divide v(˜x, y)k in R[y].

The converse is also true, since if there does not exist an αi such that αi is unequal to all

βj, then for every αi we can find a βj with αi = βj. As such we can divide v(˜x, y) by

y − αi for all i and u(˜x, y) divides v(˜x, y)k in R[y].

If we choose u as p and vk as q in the notation of 4.2.1 we get:

k _ j=0 (ak= 0 ∧ ... ∧ aj+1 = 0 ∧ aj 6= 0 ∧ (∃y)rj 6= 0) ∨ (ak = 0 ∧ ... ∧ a0= 0 ∧ (∃y)q 6= 0), (4.8)

with rj not the zero polynomial in R[y], as p did not divide q.

4.2.3 Part 3

Once again all the aido not contain y and we are left to remove the quantifier in formulas

of the following form:

(∃y)(l 6= 0),

with l a polynomial in R[ˆx, y]. This is equivalent to l not being the zero polynomial in R[y] and thus one of the coefficients in R[ˆx] must not be equal to zero.

Now ψ has been rewritten to an equivalent finite formula without quantifiers. In the next chapter we will combine above described method with the theory of Chapter 3 to give a constructive algorithm for the decidability of the equational theory of meadows.

4.3 Hilbert’s Nullstellensatz

The elimination of quantifiers has more appliances then the decidability of meadows. One of them is an alternative proof of the weak version of Hilbert’s Nullstellensatz: Theorem 4.3.1 (Hilbert’s Nullstellensatz). Let K be a field, ˆK its algebraic closure and f1(x1, ..., xn), ..., fm(x1, ..., xn) polynomials in K[x1, ..., xn]. We shall abbreviate fi(x1, ..., xn)

with fi(ˆx). Now let I be the ideal generated by these fi. Suppose that I is not equal to

the entire space K[ˆx], then there must exist ˜x in ˆK such that fi(˜x) = 0 for all i.

Proof. This proof is given in [9]. There must exist a maximal ideal M of K[ˆx] that contains I, as I is not equal to K[ˆx]. Now let us look at the image R of the canonical quotient map

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For every polynomial in M , (x1+ M, ...., xn+ M ) is an annihilator in K[ˆx]/M . Thus

there exists ˜x in F such that the equations fi(˜x) = 0 hold in F and thus also in its

algebraic closure ˆF . This means that:

(∃x1)...(∃xn)( m ^ i=1 fi(x1, .., xn)) (4.9) is true in ˆF .

Let us look at (4.9) in ˆK. Applying the quantifier elimination results in (4.9) being equivalent to a number of variables in ˆK being equal to zero (call these t1, ..., ta) and

others unequal to 0 (call these t0₁, ..., t0_a0. Since ˆK is contained in ˆF , (4.9) in ˆF is also

equivalent to t1, ..., ta being equal to zero and t01, ..., t0a0 being unequal to zero.

As (4.9) was already true in ˆF , we conclude that it is also true in ˆK, and thus there exists ˜x in ˆK such that fi(˜x) = 0 for all i.

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5 Algorithm

We shall now combine the theory from Chapter 3 and Chapter 4 to come up with a constructive algorithm to prove the decidability of the equational theory of meadows. We start by rewriting our original equation t1− t2 = 0 with variables x1, ..., xn to an

universal formula: (∀x1)...(∀xn) ((x1= 0) ∧ ... ∧ (xn= 0) ∧ (q1= 0)) ∨ (¬(x1= 0) ∧ (x2 = 0) ∧ ... ∧ (xn= 0) ∧ (q2= 0)) ∨ .. . (¬(x1= 0) ∧ ... ∧ ¬(xn= 0) ∧ (q2n = 0)). (5.1)

This form satisfies the form (∀x1)...(∀xn)(ψ(x1, ..., xn, y)) as defined in 3.2.7.

The ith line in (5.1) consists of rewriting i − 1 to the binary system and if the jth bit in the bits representation of i − 1 is a 1, then we write ¬(xj = 0) instead of (xj = 0).

Example 5.0.2. For 3 variables, the 7th line will be created as follows: 7 − 1 = 6 in the binary system is 011. So the 7th line looks like this:

((x1 = 0) ∧ ¬(x2= 0) ∧ ¬(x3 = 0) ∧ (q7 = 0)).

What remains is the definition of qi. These qi are our original equation. Since qishould

be a SMF of level 0 (see Remark 3.1.2.1), we substitute zero for all the variables in qi

that are zero and we factorize the inverses of variables that are not equal to zero out of the original equation. Since these inverses are not zero, we can remove them completely, as it has no effect on the formula being equal or unequal to zero. So from this point we no longer have any inverses in our formulas and thus all our atomic formulas are of language L.

Example 5.0.3. If our original equation is x1x−12 + x2x−13 = (x1x3+ x22)(x2x3)−1 = 0,

then q7 and q8 are equal to:

q7= x2.

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Using logical equivalences we have that (5.1) is equivalent to: ¬(∃x1)...(∃xn) (¬(x1 = 0) ∨ ... ∨ ¬(xn= 0) ∨ ¬(q1= 0)) ∧ ((x1 = 0) ∨ ¬(x2= 0) ∨ ... ∨ ¬(xn= 0) ∨ ¬(q2 = 0)) ∧ .. . ((x1 = 0) ∨ ... ∨ (xn= 0) ∨ ¬(q2n = 0)). (5.2)

Theorem 5.0.4. Let yi ∈ Rn and again write i − 1 in the binary system. Define the

jth coordinate of yi as zero if and only if the jth bit of i − 1 is a zero. Then (5.2) is

equivalent to:

2n

^

i=1

qi(fi) = 0. (5.3)

Proof. We will prove this by induction on the number of variables. If we only have one variable, then (5.2) looks like:

¬(∃x₁)(¬(x1 = 0) ∨ ¬(q1 = 0)) ∧ ((x1 = 0) ∨ ¬(q2= 0)).

Since meadows are rings, we always have that ∃x1 such that x1= 0 and ∃x1 such that

¬(x₁ = 0). So both ¬(q1 = 0) and ¬(q2 = 0) have to be false, the first one for x1 = 0

and the second one ∀x16= 0. This translates to both q1(0) = 0 and q2(y2,1) = 0 for all y2

in R\{0}. The second one is equivalent to q2 = 0, since if a polynomial is equal to zero

in infinite points it is the zero polynomial.

Assume that (5.2) is equivalent to (5.3) for n variables and look at the case that we have n + 1 variables. Then (5.2) looks like:

¬(∃x₁)...(∃xn+1) (¬(x1= 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q1= 0)) ∧ ((x1= 0) ∨ ¬(x2 = 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q2 = 0)) ∧ .. . ((x1= 0) ∨ ... ∨ (xn+1= 0) ∨ ¬(q2n+1 = 0)).

Again we have that there exist x1 such that x1= 0 and x1 such that ¬(x1 = 0). We can

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x1 = 0 is true: (∀y1 ∈ R\{0})(¬(∃x2)...(∃xn+1) ((¬(x2= 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q2(y2,1, x2, ..., xn+1) = 0)) ∧ ((x2 = 0) ∨ ¬(x3= 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q4(y4,1, x2, ..., xn+1) = 0)) ∧ .. . ((x2 = 0) ∨ ... ∨ (xn+1= 0) ∨ ¬(q2n+1(y₂n+1_,1, x₂, ..., x_n+1) = 0))) ∨ ((¬(x2= 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q1(0, x2, ..., xn+1) = 0)) ∧ ((x2 = 0) ∨ ¬(x3= 0) ∨ ... ∨ ¬(xn+1= 0) ∨ ¬(q3(0, x2, ..., xn+1) = 0)) ∧ .. . ((x2 = 0) ∨ ... ∨ (xn+1= 0) ∨ ¬(q2n+1₋₁(0, x₂, ..., x_n+1) = 0)))).

Since R has infinite elements, the formula

(∀yi,j ∈ R\{0})(qi(x1, ..., xj−1, yi,j, xj+1, ..., xn) = 0)

is equivalent to qi being the zero polynomial in R[x1, ..., xj−1, xj+1, ..., xn].

We now only have n variables left and we can apply the induction hypothesis and de-Morgan’s law to get (5.3).

To decide whether (5.3) is true, we have to check whether 2n polynomials are the zero polynomial in n variables (note that the degree of these polynomials does not have to be larger than zero in all the variables). This can be easily done by checking all the coefficients of the polynomial. The polynomial is only the zero polynomial if all these coefficients are zero.

The degree d of this polynomial is equal to the sum of the two parts of the original equation after factoring out the inverses:

Example 5.0.5. If our original equation is x1x−12 + x2x−13 = 0, then factoring out the

inverses will gives us (x1x3+ x22)(x2x3)−1 = 0. Thus the degree of our polynomials qi is

at most 4.

The number of coefficients of our polynomials qi is then at most d+n_n .

Recall that for the binomial coefficient we have the following equalities: Since we have to check the coefficients of all 2n polynomials, this gives us a complexity of O( d+n_n 2n) for our decidability algorithm.

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6 Concluding remarks

It was already known that the equational theory of meadows is decidable, but the prove was an existential one. Though most theorems in mathematics are proved in such manner, it does not immediately lead to an obvious practical application. In this thesis we have created an algorithm that can check the validity of every equation in a meadows in a finite number of steps and thus provided a constructive prove. While providing the mathematical foundation for the steps that our algorithm takes and showing these steps in Chapter 5, we have not actually implemented this algorithm. A logical step would be to implement this algorithm. In theoretical computer science however, this algorithm is certainly not needed to show that meadows are a suitable substitution for fields in ATP. The existential prove is sufficient to show that using the axioms in Σ we can verify any equality in meadows. Apart from that, the complexity of this algorithm however is exponential and thus the number of steps taken can greatly increase when the size of our input (degree of the equation) increases.

There are still a great deal of open questions regarding meadows. For instance, in this thesis we have shown that the equational theory of the class of all meadows is decidability. This does not however answer the question whether the equational theory of the meadow of rational numbers is decidable. An interesting subject for research would be to try to give an algorithm to prove that the equational theory of the meadow of rational numbers is decidable. As most programming languages already support rational number, implementation of this algorithm could result in an useful tool to prove certain mathematical theorems in the field of rational numbers.

I would like to thank my supervisors, dhr. dr. G. Regts and mw. dr. I. Bethke for supervising me with this thesis. Their patience, excellent guidance and feedback helped me in completing this project and thus also my bachelors.

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Bibliography

[1] Hiroakira Ono. “Equational theories and universal theories of fields”. In: Journal of the Mathematical Society of Japan 35.2 (1983), pp. 289–306.

[2] Jan A Bergstra, Inge Bethke, and Alban Ponse. “Cancellation Meadows: A Generic Basis Theorem and Some Applications†”. In: The Computer Journal (2012), bxs028. [3] David Marker. Model theory: an introduction. Vol. 217. Springer Science & Business

Media, 2006.

[4] Jan A Bergstra. “Elementary algebraic specifications of the rational function field”. In: Conference on Computability in Europe. Springer. 2006, pp. 40–54.

[5] Inge Bethke, Piet Rodenburg, and Arjen Sevenster. “The structure of finite meadows”. In: arXiv preprint arXiv:0903.1196 (2009).

[6] Hanamantagouda P Sankappanavar and Stanley Burris. “A course in universal algebra”. In: Graduate Texts Math 78 (1981).

[7] Joachim Lambek. Lectures on rings and modules. Vol. 283. American Mathematical Soc., 2009.

[8] Tsit-Yuen Lam. Exercises in classical ring theory. Springer Science & Business Media, 2006.

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7 Populaire samenvatting

De opkomst van computers heeft de wereld de laatste jaren flink veranderd. Elk huishouden maakt veelvuldig gebruik van de rekenkracht van deze machines. Deze rekenkracht wordt ook door onderzoekers gebruikt om hun werk te vergemakkelijken. Het vinden en bewijzen van stellingen is bijna altijd weggelegd voor de meest intelligente individuen van hun tijd. Namen als Pythagoras en Galois zijn bij de meesten wel bekend. Tegenwoordig kunnen wiskundigen computers gebruiken om hen een hand te helpen bij het bewijzen of ontkrachten van wiskunden stellingen. Door de computer een aantal regels te geven waar deze zich aan moet houden, kan de computer vervolgens zelf verder ’redeneren’ om hieruit nieuwe dingen af te leiden, in de hoop iets wat nog niet bekend is.

Zulke regels worden in de wiskunde ’axiomas’ genoemd.

Een vakgebied van de wiskunde is de universele algebra. Hierin worden algebraische structuren bekeken die voldoen aan een verzameling axiomas. Een voorbeeld van zo een algebraische structuur is een lichaam. In de wiskunde is een lichaam een verzameling getallen waarop er mag worden opgeteld, afgetrokken, vermenigvuldigd en gedeeld. De re¨ele getallen zijn een welbekend voorbeeld van een lichaam. Een groot deel axioma’s van deze lichamen zullen intu¨ıtief zijn voor de meeste mensen vanwege de re¨ele getallen:

a + b = b + a a ∗ b = b ∗ a a + 0 = 0.

Voor het delen van getallen zal ook de regel ’delen door nul is flauwekul’ de meeste mensen wel bekend in de oren klinken. In de wiskunde is dit opgeschreven door delen door nul niet te defini¨eren. Nu is deze uitzondering vervelend voor computers, omdat ze nu elke keer moeten controleren of ze wel mogen delen.

In dit verslag bekijken we een uitbreiding van de lichamen waar we delen net iets anders definieren zodat delen door nul geen problemen meer geeft:

x ∗ x x = x 1 1 x = x.

We zullen een algoritme cre¨eren dat kan controleren of een vergelijking in deze nieuwe algebra¨ısche structuur geldt of niet geldt. Wanneer deze vergelijkingen variabelen bevat, kan dit controleren vrij moeilijk worden. Algebra¨ısche structuren hoeven namelijk niet uit eindig veel elementen te bestaan, denk maar aan de re¨ele getallen. Met behulp van wiskundige eigenschap echter, kan ons algoritme deze control toch uitvoeren in eindige tijd, zelfs bij vergelijkingen met variabelen in algebra¨ısche structuren met oneindig elementen.

Decidability of the equational theory of meadows