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Definition: Magma is a couple $(A, \bot)$ with $\bot: (A \times A) \to A$

$\bot: (A \times A) \to A$ for $ \begin{cases} \bot \text{ is relation} \\ \forall x,y,z : \big ((x,y) \in \bot \wedge (x,z) \in \bot \to y=z) \big ) \\ \operatorname{dom}(\bot)=(A\times A) \\ \operatorname{cod}(\bot)\subseteq A \end{cases} $, namely $\bot $ is function from $(A\times A) $ to $A $ (binary operation on $A$)

Question to proof: $$(A, \bot) \text{ is Magma}\to \forall x,y \in A:\big ((x \bot y) \in A\big ) $$

Proof: let be $x,y\in A$ and $(x,y)$, naturally $(x,y) \in (A \times A)$. But $(A \times A)=\operatorname{dom}(\bot) $ therefore $\exists z :\bigg (\big ((x,y),z\big ) \in \bot \bigg)$ with $\bot\subseteq \big(\operatorname{dom}(\bot)\times \operatorname{cod}(\bot)\big)$, it means $(x \bot y):=\bot \big ((x,y)\big )=z \in \operatorname{cod}(\bot)$, but $\operatorname{cod}(\bot) \subseteq A $ by def. von $ \bot $, therefore $(x \bot y)\in A$

Is it correct?

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    Proof of what statement? – 2017-02-07
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    Pardon, I edited... – 2017-02-07
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    A simple definition of magma: a set with a binary operation. Apart from notation, this means what you want to prove is immediate from the definition of a binary operation. – 2017-02-07
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    From https://ncatlab.org/nlab/show/magma I read "A magma (binary algebraic structure) is a set equipped with a binary operation on it"... and not what I want to prove! But ok, wiki uses another def.. – 2017-02-07
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    But that *is* what you want to prove. Given $x,y \in A$ where $R$ denotes the binary operation on $A$, then $xRy \in A.$ To make this like your statement change the symbol $R$ to the upside down T. – 2017-02-07
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    Pardon, I don't understand your question... therefore I proved correct? – 2017-02-07
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53157/discussion-between-coffeemath-and-mle). – 2017-02-07

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