It seems obvious that 2 + 2 ≠ 5…

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

\[0 \in \mathbb{N} \label{axiom0} \\ \forall x \in \mathbb{N}, S(x) \in \mathbb{N} \\ \forall x, y \in \mathbb{N}, x = y \Leftrightarrow S(x) = S(y) \\ \forall x \in \mathbb{N}, \neg (S(x) = 0) \\ \forall x \in \mathbb{N}, x + 0 = x \\ \forall x, y \in \mathbb{N}, x + S(y) = S(x + y)\]

Which is seems more complicated than it is.

• “∀” means “for all”.
• “∈” means “element of” “belongs to” or “in”.
• “ℕ” is a symbol for the natural numbers (0, 1, 2, 3 … etc.).
• “¬” means “not”.
• “⇔” means “if and only if”.

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:

\[S(S(0)) + S(S(0)) = S(S(S(S(S(0))))) \Rightarrow \\ S(S(S(0)) + S(0)) = S(S(S(S(S(0))))) \Rightarrow \\ S(S(S(S(0)) + 0)) = S(S(S(S(S(0))))) \Rightarrow \\ S(S(S(S(0)))) = S(S(S(S(S(0)))))\]

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

\[S(S(S(S(0)))) = S(S(S(S(S(0))))) \Rightarrow \\ S(S(S(0))) = S(S(S(S(0)))) \Rightarrow \\ S(S(0)) = S(S(S(0))) \Rightarrow \\ S(0) = S(S(0)) \Rightarrow \\ 0 = S(0)\]

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

\[0 = S(0) \Rightarrow \perp\]

Therefore:

\[2 + 2 = 5 \Rightarrow \perp\]

or:

\[2 + 2 \neq 5\]