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Let universe be natural numbers:
$\mathbb{N}=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...\}$
Let this be Addition set whose elements are 3-element arrays, such that $\langle x, y, x+y \rangle$.
$\mathbb{A}=\{\langle 1, 1, 2 \rangle\, \langle 1, 2, 3 \rangle, \langle 2, 1, 3 \rangle, \langle 2, 2, 4 \rangle, \langle 1, 3, 4 \rangle, \langle 3, 1, 4 \rangle, \langle 2, 3, 5 \rangle, ...\}$
Can the following be the definition of addition in the language of predicate logic (note, $a(x, y)=x+y$):
$\forall x \forall y \exists z (a(x, y)=z \land \mathbb{A}xyz)$? Is it a problem that this formula says that there exists $z$, but it does not say that there is only one $z$?

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  • $\begingroup$ To be a definition ,you would have to know existence and uniqueness .However if you want a predicate logic for addition you introduce a 2 place function symbol a(,) .This automatically has uniqueness .So your existence condition is essentially the introduction of a new predicate for your system . So essentially you are ok. The predicate paraphrased is is really :for every x and y there is a z so that Axyz. $\endgroup$ – StuartMN Jan 28 '18 at 21:19
  • $\begingroup$ My previous comment is not adequate as the answer of A. Blass indicates .But how does the predicate Axyz get into the predicate logic $\endgroup$ – StuartMN Jan 28 '18 at 22:21
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A definition of a $k$-place function symbol $f$ should have the form $(\forall x_1)\dots(\forall x_k)(\forall y)\,[f(x_1,\dots,x_k)=y\iff\phi]$ where $\phi$ is in the original vocabulary. In your situation, the definition should read (with your choice of variables) $$ (\forall x)((\forall y)(\forall z)[a(x,y)=z\iff \mathbb Axyz]. $$

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