Term Algebras and Equational Reasoning

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Delia KESNER IRIF, CNRS et Universit ´e Paris [email protected] www.irif.fr/˜kesner

Term Algebras and Equational Reasoning

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Signatures and finite terms

Σ: Set offunction symbolshaving anarityn∈IN. X: Set ofvariables.

T(X,Σ): Set oftermsoverXandΣ:

x∈ X x∈ T(X,Σ)

t1, . . . ,tn∈ T(X,Σ) fhas arityn∈Σ f(t1, . . . ,tn)∈ T(X,Σ)

We noteVar(t)the set of variables of the termt. A termtisclosedifVar(t)=∅.

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Signatures and finite terms

Σ: Set offunction symbolshaving anarityn∈IN. X: Set ofvariables.

T(X,Σ): Set oftermsoverXandΣ:

x∈ X x∈ T(X,Σ)

t1, . . . ,tn∈ T(X,Σ) fhas arityn∈Σ f(t1, . . . ,tn)∈ T(X,Σ)

We noteVar(t)the set of variables of the termt. A termtisclosedifVar(t)=∅.

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Example Signature:

b: arity0 s: arity1

+: arity2 eq: arity2

true: arity0 f alse: arity0 Terms:

eq(s(x+b),b) eq(true,f alse) eq(s(x+b),f alse) closed

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Positions of terms

IN^∗:positionsoverIN

Λ∈IN^∗

i∈INandp∈IN^∗ ip∈IN^∗

To simplify the notation, we usually write a position by omitting the symbolΛ^,^e.g.2 denotes the position2Λ.

Pos(t):positions of a termt

Λ∈Pos(t)

p∈Pos(ti)and1≤i≤n ip∈Pos(f(t1, . . . ,tn)) Example

Pos(f(g(a,h(b)),x,c))={Λ,1,2,3,11,12,121}

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Positions of terms

Λ∈IN^∗

Λ∈Pos(t)

Pos(f(g(a,h(b)),x,c))={Λ,1,2,3,11,12,121}

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Positions of terms

Λ∈IN^∗

Λ∈Pos(t)

Pos(f(g(a,h(b)),x,c))={Λ,1,2,3,11,12,121}

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The relation ≤

_{pre f}

over positions

Concatenationof positions:

( Λ.q = q (ip).q = i(p.q)

Comparing positions: p≤_{pre f} qiff∃r∈IN^∗p.r=q Example

1 ≤_{pre f} 1211 ”1is smaller than1211”

231 ≥_{pre f} 23 ”231is greater than23”

12 ./ 2 ”12is parallel to2”(12pre f 2 & 2pre f 12)

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The relation ≤

_{pre f}

over positions

( Λ.q = q (ip).q = i(p.q)

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The relation ≤

_{pre f}

over positions

( Λ.q = q (ip).q = i(p.q)

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Sub-terms

vEt:vis asubterm/subtreeoft:

tEt

vEti

vEf(t1, . . . ,tn)

vEt:vis a subterm oft. v/t:vis astrictsubterm oft. S T(t): All the subterms oft.

Example

g(x,y)/f(g(x,y),a)anda/f(g(x,y),a)butf(x,a)6 f(g(x,y),a).

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Sub-terms

tEt

vEti

vEf(t1, . . . ,tn)

Example

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Sub-terms

tEt

vEti

vEf(t1, . . . ,tn)

Example

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Sub-term X at position Y

t|_p:subterm oftat positionp

t|_Λ=t

ti|_q=v f(t1, . . . ,tn)|iq=v

Example

f(g(a,h(b)),x,c)|11=abutf(g(a,h(b)),x,c)|21is not defined.

Exercise :Show thatp.q∈Pos(t)impliest|_p.q=(t|_p)|_q_.

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Sub-term X at position Y

t|_Λ=t

ti|_q=v f(t1, . . . ,tn)|iq=v

Example

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Sub-term X at position Y

t|_Λ=t

ti|_q=v f(t1, . . . ,tn)|iq=v

Example

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Replacement

t[p//v]:replacementof the subtermt|pby the termv t[Λ//v]=v

f(t1, . . . ,tn)[ip//v]=f(t1, . . . ,ti[p//v], . . . ,tn)

Other notations:t[v]port[v]ifpis clear from the context.

Example

f(h(x,y),a)[12//b]= f(h(x,b),a)and f(h(x,y),a)[1//b]= f(b,a).

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Replacement

Example

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Replacement

Example

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Exercise :Show the following properties:

Ifp∈Pos(s)andq∈Pos(t), then(s[t]p)|p.q=t|qand(s[t]p)[r]p.q=s[t[r]q]p. Ifp.q∈Pos(s), then(s[t]p.q)|p=(s|p)[t]qand(s[t]p.q)[r]p=s[r]p.

Ifp,q∈Pos(s)andp./q, then(s[t]p)|q=s|qand(s[t]p)[r]q=(s[r]q)[t]p.

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Σ ^-algebras

AΣ-algebraAis defined by two ingredients:

A non-emptydomainA.

Aninterpretationfunctionf^A:Aⁿ7→A, for each f/n∈Σ^. We writeA=hA,{f^A:Aⁿ7→A|f/n∈Σ}i.

Example

LetΣ ={b/0,s/1,p/2}. We give three differentΣ-algebras:

1 Ais the setIN,bÂ=0,sÂ:n7→n+1andpÂ: (n,m)7→n+m.

2 Ais the setZ^,bÂ=−5,sÂ:n7→n∗13andpÂ: (n,m)7→n∗m.

3 Syntactic Algebra:Ais the set of all the terms overXandΣsuch thatbÂ=b, sÂ:t7→s(t)andpÂ: (t,u)7→p(t,u).

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Σ ^-algebras

A non-emptydomainA.

Example

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Σ ^-algebras

A non-emptydomainA.

Example

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Σ ^-algebras

A non-emptydomainA.

Example

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Valuations

We use valuations to interpret terms in aΣ-algebra.

LetAbe aΣ-algebra and letXbe a set of variables.

AA-valuationis an applicationσ:X → A.

Definition

Given aA-valuationσ:X → A, we define a functionbσ:T(X,Σ)→ Ainterpreting arbitrary terms into theΣ-algebraAas follows:

bσ(x) = σ(x)

bσ(f(t1, . . . ,tn)) = f^A(bσ(t1), . . . ,bσ(tn))

Remark: By abuse of notation, we usually do not distinguish thevaluationσfrom the morphismbσ.

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Valuations

Definition

bσ(x) = σ(x)

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Valuations

Definition

bσ(x) = σ(x)

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Substitutions are valuations

Asubstitutionθis valuation fromXto the syntactic algebra (the set of all the terms).

Finitesubstitutions (having a finite domain) are denotedθ={x1/t1, . . . ,xn/tn}. Arenamingis an isomorphic substitution.

Example

θ1={x/y,y/x}andθ2={x/y,y/z,z/w}are renamings.

Givent= f(x,g(y))andθ={x/g(a),y/f(x,x)}we haveθ(t)= f(g(a),g(f(x,x))).

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Substitutions are valuations

Example

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Substitutions are valuations

Example

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Substitutions are valuations

Example

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Substitutions are valuations

Example

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Congruence

The symbol f∈Σismonotonicw.r.t the relationRiff aiR biimplies f^A(a1, . . . ,ai, . . . ,an)R f^A(a1, . . . ,bi, . . . ,an).

Acongruence∼is an equivalence relation (reflexive, symmetric, transitive) for a Σ-algebraAiff every symbol f∈Σis monotonic w.r.t.∼.

Example

∼={(x,y)|4dividesx−y}is a congruence.

Notation :A/∼is the set of equivalence classes of aAmodulo the congruence∼. We write[e]∼to denote the equivalent class of the elemente∈ Aw.r.t. the congruence∼.

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Congruence

Example

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Congruence

Example

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The quotient algebra

Given aΣ-algebraAwith domainA, thequotient algebraA^∼overAis defined by:

The domain isA/∼

The Interpretations aref^A^∼([a1], . . . ,[an])=[f^A(a1, . . . ,an)].

Example

LetΣ ={b/0,suc/1,pred/1,add/2}and letAbe theΣ-algebra given byA=Z^,bÂ=0, sucÂ(n)=n+1,predÂ(n)=n−1andaddÂ(n,m)=n+m.

Let∼={(x,y)|4dividesx−y}. We haveA/∼={[0],[1],[2],[3]}, where

[0] = {. . . ,−4,0,4,8, . . .} [1] = {. . . ,−3,1,5,9, . . .}

[2] = {. . . ,−6,−2,2,6,10, . . .} [3] = {. . . ,−5,−1,3,7, . . .}

We havebÂ^∼=[0],sucÂ^∼([n])=[n+1],predÂ^∼([n])=[n−1]and addÂ^∼([n],[m])=[n+m].

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The quotient algebra

Given aΣ-algebraAwith domainA, thequotient algebraA^∼overAis defined by:

The domain isA/∼

The Interpretations aref^A^∼([a1], . . . ,[an])=[f^A(a1, . . . ,an)].

Example

LetΣ ={b/0,suc/1,pred/1,add/2}and letAbe theΣ-algebra given byA=Z^,bÂ=0, sucÂ(n)=n+1,predÂ(n)=n−1andaddÂ(n,m)=n+m.

Let∼={(x,y)|4dividesx−y}. We haveA/∼={[0],[1],[2],[3]}, where

[0] = {. . . ,−4,0,4,8, . . .} [1] = {. . . ,−3,1,5,9, . . .}

[2] = {. . . ,−6,−2,2,6,10, . . .} [3] = {. . . ,−5,−1,3,7, . . .}

We havebÂ^∼=[0],sucÂ^∼([n])=[n+1],predÂ^∼([n])=[n−1]and addÂ^∼([n],[m])=[n+m].

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A -valuations vs A

^∼

-valuations

Given aA-valuationσ, we can always constructtheA^∼-valuationτgiven by τ(x)=[σ(x)]∼. We usually writeτ=σ∼.

Given aA^∼-valuationτ, we can always constructaA-valuationσs.t.

τ(x)=[σ(x)]∼. This means that we choose an elementein the equivalence class τ(x)and then we defineσ(x)=e.

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Homomorphism, endomorphism, isomorphism

LetA_andB_{be two}Σ-algebras. Amorphismis a functionΦ:A → Bs.t. for alln≥0, for all f/n∈Σand for alla1, . . . ,an∈Awe have

Φ(f^A(a1, . . . ,an))=f^B(Φ(a1), . . . ,Φ(an))

Anendomorphismover aΣ-algebraAis a morphism fromAto itself. Anisomorphismis a bijective morphism.

Theorem

For everyA-valuationσ:X → A, there is aunique morphismbσ:T(X,Σ)→ A, s.t.

bσ(x) = σ(x)

bσis the function that we have defined so far to interpret arbitrary terms in aΣ^-algebra.

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Homomorphism, endomorphism, isomorphism

Φ(f^A(a1, . . . ,an))=f^B(Φ(a1), . . . ,Φ(an))

Theorem

bσ(x) = σ(x)

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Homomorphism, endomorphism, isomorphism

Φ(f^A(a1, . . . ,an))=f^B(Φ(a1), . . . ,Φ(an))

Theorem

bσ(x) = σ(x)

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Homomorphism, endomorphism, isomorphism

Φ(f^A(a1, . . . ,an))=f^B(Φ(a1), . . . ,Φ(an))

Theorem

bσ(x) = σ(x)

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Exercise :LetΣ ={a/0,s/1,g/1,h/2}. Define aΣ-algebraBand a morphismΦbetween the syntacticΣ-algebra andB.

LetB_{be the}Σ-algebradefined by:

The domainB={n∈IN|n≥2}

The interpretations:

a^B = 2 g^B(t) = t+1

s^B(t) = t h^B(u,t) = u+t LetΦbe the following function:

Φ(a) = 2 Φ(g(t)) = Φ(t)+1 Φ(s(t)) = Φ(t) Φ(h(u,t)) = Φ(u)+ Φ(t) One verifiesΦ(f^A(t1, . . . ,tn))=f^B(Φ(t1), . . . ,Φ(tn))for everyf/ninΣ^.

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a^B = 2 g^B(t) = t+1

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a^B = 2 g^B(t) = t+1

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a^B = 2 g^B(t) = t+1

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Semantical equational reasoning

AΣ-equationis a pair of terms denotedst.

Definition

LetAbe aΣ-algebra.

Ais amodelofan equationst, writtenA |=st, iff the equalitybσ(s)=bσ(t) holds for everyA-valuationσ.

Ais amodelof aset ofΣ-equationsE, writtenA |=E, iffAis a model of every equation inE.

Example

The firstΣ-algebra on slide (Σ-algebras) is a model of the equationp(x,y)p(y,x). What about the second and the third?

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Semantical equational reasoning

Definition

LetAbe aΣ-algebra.

Example

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Semantical equational reasoning

Definition

LetAbe aΣ-algebra.

Example

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Semantical equational reasoning

Definition

LetAbe aΣ-algebra.

Example

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Semantical equational reasoning

Definition

LetAbe aΣ-algebra.

Example

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Properties of semantical equational reasoning

A |=ss.

IfA |=st, thenA |=ts.

IfA |=standA |=tu, thenA |=su.

IfA |=st, then∀u∀p∈Pos(u),A |=u[s]_pu[t]_p.

Proof.

By induction onu.

IfA |=st, thenA |=θ(s)θ(t), for every substitutionθ.

Proof.

We want to provebσ(θ(s))=bσ(θ(t))for everyA-valuationσ, so that let us take an arbitrary A-valuationσ. Letτσbe theA-valuation given byτσ(x)=bσ(θ(x)). We first show by induction onuthatτbσ(u)=bσ(θ(u))(easy). Now,A |=stimpliesbτ(s)=bτ(t)for every A-valuationτ, so that in particular forτσ. Thus,τbσ(s)=bσ(θ(s))=bσ(θ(t))=τbσ(t).

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Properties of semantical equational reasoning

A |=ss.

Proof.

By induction onu.

IfA |=st, thenA |=θ(s)θ(t), for every substitutionθ.

Proof.

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Properties of semantical equational reasoning

A |=ss.

Proof.

By induction onu.

IfA |=st, thenA |=θ(s)θ(t), for every substitutionθ. Proof.

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Properties of semantical equational reasoning

A |=ss.

Proof.

By induction onu.

IfA |=st, thenA |=θ(s)θ(t), for every substitutionθ. Proof.

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Semantic equational consequence

Definition

The equationstis asemantic consequenceof a set of equationsE, writtenE |=st, iff every model ofEis also a model ofstiff for everyΣ-algebraA,A |=Eimplies A |=st.

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Syntactic rules for equational reasoning

st∈ E

st ^(axiom) ss (reflexivity)

st

ts ^(symmetry)

st tu

su (transitivity)

st

θ(s)θ(t) (substitution) st u[s]pu[t]p

(context)

Derivationofstfrom the setE(which gives a tree) is denoted byE `st.

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Syntactic rules for equational reasoning

st∈ E

st

ts ^(symmetry)

st tu

su (transitivity)

st

(context)

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Syntactic rules for equational reasoning

st∈ E

st

ts ^(symmetry)

st tu

su (transitivity)

st

(context)

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Example

LetE={0+zz,s(y)+xs(y+x)}. Let us use the notations3=s(s(s(0)))and 4=s(s(s(s(0)))).

We derives(0)+34fromEas follows:

s(y)+xs(y+x)∈ E

(axiom)

s(y)+xs(y+x)

(substitution)

s(0)+3s(0+3)

0+zz∈ E

(axiom)

0+zz

(substitution)

0+33

(context)

s(0+3)s(3)

(transitivity)

s(0)+3s(3)

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The relation ↔

^∗_E

st∈ E θ(s)↔Eθ(t)

st∈ E θ(t)↔Eθ(s)

s↔_Et u[s]p↔Eu[t]p

↔^∗_Eis the reflexive-transitive closure of↔E.

Exercise :

1 ↔^∗_Eis stable by substitution.

2 ↔^∗_Eis an equivalence relation.

3 ↔^∗_Eis a congruence overT(X,Σ).

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The relation ↔

^∗_E

Exercise :

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The relation ↔

^∗_E

Exercise :

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Properties of ↔

^∗_E

Exercise :E `stif and only ifs↔^∗_Et.

Proof.

Left-Right: by induction on derivation ofE `st. Right-Left: by induction on the length of↔^∗_E.

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Properties of ↔

^∗_E

Proof.

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Properties of ↔

^∗_E

Proof.

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Substitution Lemma

Lemma

Letσbe anyA-valuation, and letτbe theA^∼-valuation defined byτ:x7→[σ(x)]∼. Then bτ(t)=[bσ(t)]∼for every termt.

Proof.

By induction ont.

Fort=xthe property holds by definition.

Fort=f(t1, ..,tn), we have

bτ(f(t1, ..,tn)) = f^A^∼(bτ(t1), . . . ,bτ(tn)) =h.r

f^A^∼([bσ(t1)]∼, . . . ,[bσ(tn)]∼) =de f [f^A(bσ(t1), . . . ,bσ(tn))]∼ =

[bσ(t)]∼ =

A particular case of this Lemma is given whenσis a substitution andτa T(X,Σ)^↔^∗^E-valuation.

An even more particular case is whenσis theidsubstitution so thatbτ(t)=[t]↔^∗ E.

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Substitution Lemma

Lemma

Proof.

By induction ont.

[bσ(t)]∼ =

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Substitution Lemma

Lemma

Proof.

By induction ont.

[bσ(t)]∼ =

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T (X, Σ )

^↔^∗^E

as a model

Theorem

T(X,Σ)^↔^∗^Eas a model ofE.

Proof.

Take any equationst∈ E. Take anyT(X,Σ)^↔^∗^E-valuationτ. By previous remark we can construt aT(X,Σ)-valuationσ(which is a substitution in this case) such that

τ(x)=[σ(x)]↔^∗

E. The substitution lemma guarantees thatbτ(t)=[bσ(t)]↔^∗

Efor every termt. Now,st∈ Eimplies (by def)

E `stiff (previous exercise)s↔^∗_Etimplies (↔^∗_Eis stable by substitution) bσ(s)↔^∗_Ebσ(t)iff (by def)[bσ(s)]↔^∗

E=[bσ(t)]↔^∗

Eiff (Subst. Lemma)bτ(s)=bτ(t).

Since anyT(X,Σ)^↔^∗^E-valuationτsatifies the equationsE:T(X,Σ)^↔^∗^Eis a model ofE.

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T (X, Σ )

^↔^∗^E

as a model

Theorem

T(X,Σ)^↔^∗^Eas a model ofE.

Proof.

Take any equationst∈ E. Take anyT(X,Σ)^↔^∗^E-valuationτ. By previous remark we can construt aT(X,Σ)-valuationσ(which is a substitution in this case) such that

τ(x)=[σ(x)]↔^∗

E. The substitution lemma guarantees thatbτ(t)=[bσ(t)]↔^∗

Efor every termt. Now,st∈ Eimplies (by def)

E `stiff (previous exercise)s↔^∗_Etimplies (↔^∗_Eis stable by substitution) bσ(s)↔^∗_Ebσ(t)iff (by def)[bσ(s)]↔^∗

E=[bσ(t)]↔^∗

Eiff (Subst. Lemma)bτ(s)=bτ(t).

Since anyT(X,Σ)^↔^∗^E-valuationτsatifies the equationsE:T(X,Σ)^↔^∗^Eis a model ofE.

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Birkhoff’s Theorem (1933)

LetEbe a set ofΣ-equations.

(Soundness)IfE `st, thenE |=st.

Proof.

By induction onE `stusing the properties of semantical equational reasoning.

(Completeness)IfE |=st, thenE `st.

Proof.

IfE |=st, then every model ofEis a a model ofst. SinceT(X,Σ)^↔^∗^Eis a model ofE (previous theorem), then in particularT(X,Σ)^↔^∗^E|=st.

This means thatbτ(s)=bτ(t)for everyT(X,Σ)^↔^∗^E-valuationτ, so in particular for τid:x7→[id(x)]↔^∗

E. Thus,τcid(s)=cτid(t). By the Substitution Lemmaτcid(s)=[bid(s)]↔^∗

Eandcτid(t)=[bid(t)]↔^∗ E, thus [bid(s)]↔^∗

E=[s]↔^∗ E=[t]↔^∗

E=[bid(t)]↔^∗

Ewhich meanss↔^∗_Et. We concludeE `st.

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Birkhoff’s Theorem (1933)

Proof.

E=[s]↔^∗ E=[t]↔^∗

E=[bid(t)]↔^∗

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Birkhoff’s Theorem (1933)

Proof.

E=[s]↔^∗ E=[t]↔^∗

E=[bid(t)]↔^∗

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Birkhoff’s Theorem (1933)

Proof.

E=[s]↔^∗ E=[t]↔^∗

E=[bid(t)]↔^∗

Term Algebras and Equational Reasoning