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Written By: Quarkoala

May 17, 2020

FREE WRITING

"So class, what's 1+1?"

"2-"

"PROVE IT!"

...

...

...

Okay, let's do this. First of all, we're working in the natural numbers (the infinite set of numbers {0,1,2,3,4,5,...}), because this is at most a kindergarten class, and possibly pre-kindergarten. We don't need to know the precise recursive definition of the natural numbers, just that it exists, and that it says that every natural number n has a uniques successor, S(n), and that one of the natural numbers is 0. We will need to know the recursive definition of addition:

(i) For any natural number n, n+0=n.

(ii) For any natural numbers m and n, S(m+n)=m+S(n).

We also need to know what 1 and 2*are:*

1=S(0)

2=S(1)=S(S(0))

3=S(2)=S(S(S(0)))

4=S(3)=S(S(S(S(0))))

You get the pattern.

Oh, and we should know what a theorem is: a statement that can be proven rigorously from other theorems and axioms. Axioms are also statements, but can't be proven, and we just assume them to be true.

All right, now we can do the proof. When it is complete, we will write QED.

**Theorem **1+1=2.

**Proof **1+1=S(0)+S(0)=S(S(0)+0)=S(S(0))=2.

QED

...

...

...

"But I wanted a completely formal and rigorous*first order logic *proof."

...

...

...

Okay, fine! First order logic it is!

We'll need to know a bit more before doing this proof. We write "implies" with the symbol ⇒, "and" with the symbol ^, and "or" with the symbol v. ∀ means "for all" and ∃ means "there exists." We'll need to know the properties of the identity relation, =:

Symmetry: (x=y)⇒(y=x)

Transitivity: ((x=y)^(y=z))⇒(x=z)

Reflexive: (∀x)(x=x)

We will abbreviate all of these properties as Id.

The defining axiom of a function (which is a type of relation) f is:

(∀w)((∀x)((∀y)((∀z)(((wfx^yfz)^w=y)⇒x=z))))

We define f(x) to be such that (x)f(f(x)), and by the above definition, f(x) is unique (that's really what it's saying, it just ends up looking really complicated).

We will take it for granted that the function f defined by f(x)=x+a where a is a constant natural number is indeed a function, and we will not write anything when we use this fact (you'll see what I mean when you see the proof). We will also take for granted that S is a function, the successor function.

For anybody who is reading this who is really picky, no, we will not be using Universal Instantiation and Generalization. For everybody who doesn't know what those are, don't worry, we're not using them.

Sheesh, that was a lot! But now we can do the proof. We will start with the statements: 1=S(0) and 2=S(S(0)). Also, we will need to use (i) and (ii) from earlier. I know I didn't explain the format of the proof, but I think you'll catch on. I'll tell you this: each line of the proof has a number, a true statement (that we started off with, or that follows from previous statements), and a list of the numbers/symbols/tags of statements that were used to prove it. The statements on each line are true statements, so we want 1+1=2 to become one of them, for then we would know it is true.

**Theorem** 1+1=2

**Proof**

1) 1=S(0)

2) 2=S(S(0))

3) 1+1=S(0)+S(0) (1)

4) S(0)+S(0)=S(S(0)+0) (ii)

5) S(0)+0=S(0) (i)

6) S(S(0)+0)=S(S(0)) (5)

7) 1+1=S(S(0)+0) (3, 4, Id.)

8) 1+1=S(S(0)) (6, 7, Id.)

9) S(S(0))=2 (2, Id.)

10) 1+1=2 (8, 9, Id.)

QED

...

...

...

"But I wanted a completely formal and rigorous*second** *order logic proof."

...

...

...

You know what, forget about it.

"2-"

"PROVE IT!"

...

...

...

Okay, let's do this. First of all, we're working in the natural numbers (the infinite set of numbers {0,1,2,3,4,5,...}), because this is at most a kindergarten class, and possibly pre-kindergarten. We don't need to know the precise recursive definition of the natural numbers, just that it exists, and that it says that every natural number n has a uniques successor, S(n), and that one of the natural numbers is 0. We will need to know the recursive definition of addition:

(i) For any natural number n, n+0=n.

(ii) For any natural numbers m and n, S(m+n)=m+S(n).

We also need to know what 1 and 2

1=S(0)

2=S(1)=S(S(0))

3=S(2)=S(S(S(0)))

4=S(3)=S(S(S(S(0))))

You get the pattern.

Oh, and we should know what a theorem is: a statement that can be proven rigorously from other theorems and axioms. Axioms are also statements, but can't be proven, and we just assume them to be true.

All right, now we can do the proof. When it is complete, we will write QED.

QED

...

...

...

"But I wanted a completely formal and rigorous

...

...

...

Okay, fine! First order logic it is!

We'll need to know a bit more before doing this proof. We write "implies" with the symbol ⇒, "and" with the symbol ^, and "or" with the symbol v. ∀ means "for all" and ∃ means "there exists." We'll need to know the properties of the identity relation, =:

Symmetry: (x=y)⇒(y=x)

Transitivity: ((x=y)^(y=z))⇒(x=z)

Reflexive: (∀x)(x=x)

We will abbreviate all of these properties as Id.

The defining axiom of a function (which is a type of relation) f is:

(∀w)((∀x)((∀y)((∀z)(((wfx^yfz)^w=y)⇒x=z))))

We define f(x) to be such that (x)f(f(x)), and by the above definition, f(x) is unique (that's really what it's saying, it just ends up looking really complicated).

We will take it for granted that the function f defined by f(x)=x+a where a is a constant natural number is indeed a function, and we will not write anything when we use this fact (you'll see what I mean when you see the proof). We will also take for granted that S is a function, the successor function.

For anybody who is reading this who is really picky, no, we will not be using Universal Instantiation and Generalization. For everybody who doesn't know what those are, don't worry, we're not using them.

Sheesh, that was a lot! But now we can do the proof. We will start with the statements: 1=S(0) and 2=S(S(0)). Also, we will need to use (i) and (ii) from earlier. I know I didn't explain the format of the proof, but I think you'll catch on. I'll tell you this: each line of the proof has a number, a true statement (that we started off with, or that follows from previous statements), and a list of the numbers/symbols/tags of statements that were used to prove it. The statements on each line are true statements, so we want 1+1=2 to become one of them, for then we would know it is true.

1) 1=S(0)

2) 2=S(S(0))

3) 1+1=S(0)+S(0) (1)

4) S(0)+S(0)=S(S(0)+0) (ii)

5) S(0)+0=S(0) (i)

6) S(S(0)+0)=S(S(0)) (5)

7) 1+1=S(S(0)+0) (3, 4, Id.)

8) 1+1=S(S(0)) (6, 7, Id.)

9) S(S(0))=2 (2, Id.)

10) 1+1=2 (8, 9, Id.)

QED

...

...

...

"But I wanted a completely formal and rigorous

...

...

...

You know what, forget about it.

That got very technical very fast. If you want, comment how much you followed, and if you enjoyed this piece or not. Also let me know if I made a mistake in any of the explanations and proofs, and if there were any typos. Nice day!

## 4 Comments

journal.scribbles

Wow, impressive. I'd like to say that I followed the whole thing but... :)

sunny.v

oh and i agree with @Cressida! restating 2 in 9 isn’t necessary but i mean on an exam i’m sure the grader would appreciate it, so it works! but for the sake of writing and shortening, perhaps cut 9 out?

sunny.v

helppp i’ve longed for a math piece of writing all my life. thank you for this! proofs are the actual bane of precalc. statements and reasoning on a t-chart? no. NO. actual war flashbacks, dude

Cressida

I followed it, but I do have a question. I haven't done many proofs so correct me if I'm wrong, but when you start discussing S(n), you state that 2= S(S(0)) (which makes sense since S(0) = 1 and 1 + 1 is 2), but if you're trying to PROVE that 1 + 1 = 2, how can you use that in your proof? I feel like just to comprehend that S(S(0))= 2, you would need to know that 1 + 1=2, which you wouldn't if you're trying to prove that very fact, if that makes sense? And it might even go against proofs to use the idea you're trying to prove as a fact/reason for why the proof is true?

Regardless, I think to shorten your proof a little bit you could get rid of step 9 in your proof and jump right to step 10, since step 9 is just a reiteration of step 2, so instead of using step 9 to back up step 10 just use step 2 instead.

Not trying to poke at your proof or anything with what I was saying before, and I hope you don't take that the wrong way, I truly did enjoy reading something math-based on this website :)