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    $\begingroup$ Nice answer. Note that axiom1 should say gcd[a, a] == a. $\endgroup$ Commented Mar 30, 2020 at 22:58
  • $\begingroup$ @Jose: Nice catch! I've included your correction... which shortened the proof, by the way. $\endgroup$ Commented Mar 31, 2020 at 2:45
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    $\begingroup$ +1. In fact, axioms 4,5, and 6 are not axioms, but the hypotheses of the theorem. I suggest hyp1, hyp 2, and hyp 3 would be exacter notations. $\endgroup$ Commented Mar 31, 2020 at 8:52
  • $\begingroup$ @user64494: Thanks for your clarifying comment. I've attempted to incorporate your suggestions. $\endgroup$ Commented Mar 31, 2020 at 17:40