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It seems like a simple question, but is actually quite hard to prove. You don't just simply say:

$a = c$ because $a = b$ and $b = c$.

Like I actually want a proof behind it, something like using formulae, arithmetic, etc.

How is this possible?

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    [The algebra tag is no longer being used, as "algebra" can mean very different things.](http://math.stackexchange.com/tags/algebra/info)2012-09-03

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In any part of mathematics where one uses equality (and I can't think of any parts where one doesn't) one admits as axioms those that express that equality is an equivalence relation. This means that you can always assume that

  • For any value $x$ one has $x=x$,
  • Whenever $x=y$ then also $y=x$,
  • Whenever $x=y$ and $y=z$ then also $x=z$.

The question you ask is the third axiom, so you can use this without needing to prove it.

Any notion of "equality" that does not satisfy these axioms should not be denoted by "$=$", as this would immediately invite erroneous arguments that do use the above properties. An example of such a relation is "approximately equal" among (for instance) real numbers, formulated using any reasonable precise definition you like.

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    The Q asks about relational property of equality and we were discussing categorical formalization of antisymmetry in pre-orders. A poset is a skeletal pre-order category.2013-02-22
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I'd just like to point out that at least in ZF set theory, a current standard foundation for mathematics, equality is defined in terms of membership and simple logic, so this becomes a theorem.

Definition:

$A = B \iff (x \in A \iff x \in B)$

Theorem (Transitivty of Equality):

$(A = B \wedge B = C) \implies A = C$

Proof:
Suppose: $A = B \wedge B = C$ Applying the definition of equality: $(x \in A \iff x \in B) \wedge (x \in B \iff x \in C)$ By transitivity of logical equivalence (which I believe is a theorem of elementary logic): $(x \in A \iff x \in C)$ Which by definition is, $A = C$

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    Logical equivalence is usually defined as an abbreviation: $x\leftrightarrow y$ means $(x\to y)\land(y\to x)$. Since $(x\to y)\land(y\to z)\implies(x\to z)$ is either an axiom or easily derivable, one can prove without difficulty $(x\leftrightarrow y)\land(y\leftrightarrow z)\implies(x\leftrightarrow z)$.2013-03-17
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This is the transitive property. Many relations depend on the transitive property, such as partial order relations and equivalence relations. You can read more about relations here http://en.wikipedia.org/wiki/Binary_relation or for example, Munkres Topology gives a nice little introduction good enough for many practical purposes.