# On Arithmetic

# 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:

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:

*0 is a natural number**For every natural number x, the number after x is also a natural number.**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**0 does not come after any natural number.**Any natural number plus zero is itself**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 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:

Then we apply our axioms of addition:

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

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

Therefore:

or:

Qed.

*Originally published at **andymac-2.github.io**.*