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as the title says I am trying to give a (nearly, but not fully formal) proof that the weak axiom of pairing (i.e. $\forall x \forall y \exists p: x \in p \wedge y \in p$) together with a suitable instance of the axiom schema of specification does imply the axiom of pairing. I haven't found a suitable instance yet, so this would be the first step to take.

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Here is a 'nearly, but not fully, formal proof' (OK, it is fully formal, but it is not fully completed):

enter image description here

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    $\begingroup$ What software is that? $\endgroup$ – J.G. Jan 12 at 18:46
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    $\begingroup$ @J.G. It's called 'Fitch' .. it comes with the book "Language, Proof, and Logic" $\endgroup$ – Bram28 Jan 12 at 19:14
  • $\begingroup$ This software seems awesome! (Though I don't fully understand the notions there.) Thank you for your reply! $\endgroup$ – Studentu Jan 13 at 18:10
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    $\begingroup$ @Studentu The 'Fitch' system is actually a fairly well known and well-used system for creating fully formal proofs. Here is a list with all the rules: math.mcgill.ca/rags/JAC/124/Rules-Strategy-b.pdf $\endgroup$ – Bram28 Jan 13 at 18:21
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    $\begingroup$ @Studentu Cool. You're welcome! :) $\endgroup$ – Bram28 Jan 15 at 1:31
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Given $x,y$, let $p$ be such that $x\in p\land y\in p$. Then $\{x,y\}=\{\,t\in p\mid t=x\lor t=y\,\}$.

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  • $\begingroup$ Thanks for your answer! $\endgroup$ – Studentu Jan 13 at 18:09
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Fix $x,y$ and take $p$ such that $x, y \in p$. Now take the formula $\Phi = ( z = x \lor z = y)$ and apply the axiom of specification on $p$.

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  • $\begingroup$ Thank you for answering! $\endgroup$ – Studentu Jan 13 at 18:09

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