Unique homomorphic extension theorem

From HandWiki

The unique homomorphic extension theorem is a result in mathematical logic which formalizes the intuition that the truth or falsity of a statement can be deduced from the truth values of its parts.[1][2][3]

The lemma

Let A be a non-empty set, X a subset of A, F a set of functions in A, and X+ the inductive closure of X under F.

Let be B any non-empty set and let G be the set of functions on B, such that there is a function d:F→G in G that maps with each function f of arity n in F the following function d(f):Bn→B in G (G cannot be a bijection).

From this lemma we can now build the concept of unique homomorphic extension.

The theorem

If X+ is a free set generated by X and F, for each function h:X→B there is a single function h^:X+→B such that:

∀x∈X,h^(x)=h(x);(1)

For each function f of arity n > 0, for each x1,…,xn∈X+n,

h^(f(x1,…,xn))=g(h^(x1),…,h^(xn)), where g=d(f)(2)

Consequence

The identities seen in (1) e (2) show that h^ is an homomorphism, specifically named the unique homomorphic extension of h. To prove the theorem, two requirements must be met: to prove that the extension (h^) exists and is unique (assuring the lack of bijections).

Proof of the theorem

We must define a sequence of functions hi:Xi→B inductively, satisfying conditions (1) and (2) restricted to Xi. For this, we define h0=h, and given hi then hi+1shall have the following graph:

{(f(x1,…,xn),g(hi(x1),…,hi(xn)))∣(x1,…,xn)∈Xin−Xi−1n,f∈F}∪graph⁡(hi) with g=d(f)

First we must be certain the graph actually has functionality, since X+ is a free set, from the lemma we have  f(x1,…,xn)∈Xi+1−Xi when (x1,…,xn)∈Xin−Xi−1n,(i≥0), so we only have to determine the functionality for the left side of the union. Knowing that the elements of G are functions(again, as defined by the lemma), the only instance where (x,y)∈graph(hi) and (x,z)∈graph(hi) for some x∈Xi+1−Xi is possible is if we have  x=f(x1,…,xm)=f′(y1,…,yn)  for some (x1,…,xm)∈Xim−Xi−1m,(y1,…,yn)∈Xin−Xi−1n and for some generators f and f′ in F.

Since f(X+m) and f′(X+n)  are disjoint when f≠f′,f(x1,…,xm)=f′(y1,…,Yn) this implies f=f′ and m=n. Being all f∈F in X+n, we must have xj=yj,∀j,1≤j≤n.

Then we have y=z=g(x1,…,xn) with g=d(f), displaying functionality.

Before moving further we must make use of a new lemma that determines the rules for partial functions, it may be written as:

 (3)Be (fn)n≥0 a sequence of partial functions fn:A→B such that fn⊆fn+1,∀n≥0. Then, g=(A,⋃graph(fn),B) is a partial function. [1]

Using (3), h^=⋃i≥0hi is a partial function. Since  dom(h^)=⋃dom(hi)=⋃Xi=X+ then h^ is total in X+.

Furthermore, it is clear from the definition of hi that h^ satisfies (1) and (2). To prove the uniqueness of h^, or any other function h′ that satisfies (1) and (2), it is enough to use a simple induction that shows h^ and h′ work for Xi,∀i≥0, and such is proved the Theorem of the Unique Homomorphic Extension.[2]

Example of a particular case

We can use the theorem of unique homomorphic extension for calculating numeric expressions over whole numbers. First, we must define the following:

A=Σ* where Σ=Variables∪{0,1,2,…,9}∪{+,−,*}∪{(,)}, where |*=Variables∪{0,…,9}

Be F={f−,f+,f*}

f:Σ*→Σw↦−w*

f:Σ*xΣ*→Σw1,w2↦w1+w2*

f:Σ*xΣ*→Σw1,w2↦w1*w2*

Be EXPR he inductive closure of X under F and beB=ℤ,G={Soma(−.−),Mult(−,−),Menos(−)}

Be d:F→G

d(f−)=menos

d(f+)=mais

d(f*)=mult

Then h^:X+→{0,1} will be a function that calculates recursively the truth-value of a proposition, and in a way, will be an extension of the function h:X→{0,1}that associates a truth-value to each atomic proposition, such that:

(1)h^(ϕ)=h(ϕ)

(2)h^((¬ϕ))=NAO(h^(ψ)) (Negation)

h^((ρ∧θ))=E(h^(ρ),h^(θ)) (AND Operator)

h^((ρ∨θ))=OU(h^(ρ),h^(θ)) (OR Operator)

h^((ρ→θ))=SEENTAO(h^(ρ),h^(θ)) (IF-THEN Operator)

References