Piano’s Fifth Axiom
This blog covers the proof of Piano’s Fifth Axiom using set theoretic definitions which are being used to develop Piano’s Axioms.
Axiom: If \(n\) and \(w\) are in \(\omega\), and if \(n^{+} = m^{+}\) then \(n = m\).
To prove the given axiom, we need to first go through two propositions,
Proposition 1. No natural number is a subset of any of it’s elements.
Proof 1: Consider the set \(S\) with the given properties, now as \(0 = \emptyset\), therefore no elements in \(0\) exists hence \(0\) is not a subset of any of it’s elements or \(0 \in S\).
Induction Step: Consider \(n \in S\), now as \(n = n\) therefore \(n \subset n\) hence \(n\) can not be an element of \(n\) i.e. \(n \notin n\) as otherwise \(n \notin S\). Considering, \(n^{+} = n \cup \{n\}\), as \(n \notin n\) therefore \(n^{+} \not\subset n\) — (1).
Now let \(n^{+} \subset x\) therefore \(n \subset x\), which gives \(x \notin n\), therefore \(n^{+}\) is not a subset of any \(x \in n\). Hence, \(n^{+}\) is not a subset of any element of \(n^{+}\) showing that \(n^{+} \in S\).
Now by mathematical induction, \(S = \omega\) (the set of natural numbers).
Reference: Naive Set Theory, Paul Halmos