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SHOCK! HORROR! My own friend Wojowu DARES to make treacherously false claims about the nature of mathematical logic itself!

In a scandalous blog post of his, he claims the existence of this number called "\(N\)" — which he also calls the "Robinson number" — that turns out to be larger than every natural number. He then goes on to show how the existence of \(N\) is derived from the axioms of Robinson's Q (which I will abbreviate as just Q from now on).

As googologists, we know this to be a load of absolute balderdash! There is no "infinity" or any kind of "end" to the natural numbers! That's the whole point of our adventure, no?!

Recall the axioms of Q:

  1. \(Sx\neq 0\)
  2. \((Sx=Sy)\Rightarrow x=y\)
  3. \(x=0\lor \exists y(Sy=x)\)
  4. \(x+0=x\)
  5. \(x+Sy=S(x+y)\)
  6. \(x\cdot 0=0\)
  7. \(x\cdot Sy=x\cdot y+x\)

where \(S\) is a unary operator and \(+,\cdot\) are binary operators. We also define \(x\leq y \Leftrightarrow \exists z: x+z=y\).

Wojowu, in a feat of dastardly metamathematical trickery, attempts to define his "\(N\)" and notes that any operation involving \(N\) results in \(N\), save for \(N \cdot 0 = 0\). A consequence is that \(SN = N\).

This is a load of baloney! We can easily show that for all \(x\), \(Sx \neq x\). To do this, we can construct a model of Q that satisfies this statement. We define \(0 =_\text{def} \{\}\), \(Sx =_\text{def} x \cup \{x\}\), and consider the model to be the set \(\{0, S0, SS0, SSS0, SSSS0, \ldots\}\). We go through the axioms to verify that they are all satisfied by the model:

  1. \(Sx \neq 0\): The definition of \(Sx\) requires that it must contain at least one element, and since 0 is the empty set, this axiom is true.
  2. \((Sx=Sy) \Rightarrow x=y\): This is a straightforward consequence of extensionality.
  3. \(x=0\lor \exists y(Sy=x)\): Every member of our model, save zero, is of the form \(Sy\), so this is true.

The remaining axioms are simply definitions and it's almost trivial to show that the model satisfies them. Notice that for none of the elements in the model is \(Sx = x\), so the notion that \(Sx \neq x\) is perfectly logical. This means that \(SN = N\) is wrong, and by the law of contrapositive \(N\) CANNOT EXIST!

I've shown to you from the axioms of Q that the existence of \(N\) is a gigantic truckload of hogwash! Bovine feces! I have EXPOSED Wojowu as a LIAR and a FAKE!

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