Proving the Equivalence of Solving Equations and Finding Functions

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tickle_monste
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I've got a proof but I'll wait a couple days to post mine to give you guys a chance to take a crack at it.

Prove that:

For all equations E(A1, ... , An), solving for Ai is equivalent to finding the function
f(A1..A(i-1),A(i+1)...An), such that when f is substituted in E in place of Ai, E reduces to an identity (i.e. 0 = 0, 1 = 1, a^2 + b^2 = c^2, etc.)

Use whatever axioms you'd like.
 
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Seems like implicit function theorem proof.
 
Not necessary.
 
I don't think this is true. If you consider the equation x_1=x_1^2 and the function f(x_2)=0 reduces to the identity 0 = 0, but misses the solution x_1=1.
 
My wording of the problem was poor, I forgot to account for multiple independent solutions. So, here it is, better worded:
Finding a solution of Q in the equation is equivalent to finding a function f, such that when f is substituted for Q in the equation, the equation will reduce to an identity.

Both of your solutions satisfy this property, no other number satisfies this property and no other number is a solution.
 
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC). Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?
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