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It seems obvious that 2 + 2 ≠ 5…

Or is it?

A rigorous proof that 2 + 2 ≠ 5 is not particularly trivial. In order to be concise, we need to begin with some definitions:

0NxN,S(x)Nx,yN,x=yS(x)=S(y)xN,¬(S(x)=0)xN,x+0=xx,yN,x+S(y)=S(x+y)

Which is seems more complicated than it is.

The only other part here that hasn’t been described is “S()”, the successor function. The successor of a number is the number after itself, for example, S(0) = 1, and S(3) = 4.

The axioms above in English for reference are as follows:

  1. 0 is a natural number
  2. For every natural number x, the number after x is also a natural number.
  3. For any two natural numbers x and y, if x = y, then S(x) = S(y) and if S(x) = S(y) then x = y
  4. 0 does not come after any natural number.
  5. Any natural number plus zero is itself
  6. For any two natural numbers x and y, x + S(y) = S(x + y)

This is sufficient to prove 2 + 2 ≠ 5. For brevity we have left out some axioms of equality.

We define 2 as S(S(0)) and we define 5 as S(S(S(S(S(0))))). To prove that 2 + 2 ≠ 5, we need to prove ¬(2 + 2 = 5). ¬(2 + 2 = 5) is equivalent to 2 + 2 = 5 ⇒ ⊥ where “⇒” means “implies” and “⊥” means “falsehood” or “contradiction”. So if we can prove that 2 + 2 = 5 implies a contradiction or falsehood, then we have our proof.

We start with our definition of 2 and 5:

S(S(0))+S(S(0))=S(S(S(S(S(0)))))

Then we apply our axioms of addition:

S(S(0))+S(S(0))=S(S(S(S(S(0)))))S(S(S(0))+S(0))=S(S(S(S(S(0)))))S(S(S(S(0))+0))=S(S(S(S(S(0)))))S(S(S(S(0))))=S(S(S(S(S(0)))))

We use axiom 3 to reduce our problem by removing one layer of successor functions each time.

S(S(S(S(0))))=S(S(S(S(S(0)))))S(S(S(0)))=S(S(S(S(0))))S(S(0))=S(S(S(0)))S(0)=S(S(0))0=S(0)

For which we already know by axiom 4 that ¬(S(0) = 0), or as we have seen before, S(0) = 0 ⇒ ⊥.

0=S(0)⇒⊥

Therefore:

2+2=5⇒⊥

or:

2+25