Empty types

(and empty sets)

\[\vdash \bot:type (F)\]

 

\[\frac{A:type}{f:\bot \to A}(E)\]

"The" empty type (falsity)

Indeed, \(P:Prop\) then \(f:\bot\to P\).  So ...

if ever anyone introduces \(e:\bot\) then \(f(e):P\).

I.e. if "falsity" (\(\bot\)) is "true" then everything is true and false.

Other empty types

• Data can only be equal by reflexive: \[\frac{x:A}{refl_x:x=_Ax}(I)\]
• Hence no evidence for \(0=1=S(0)\) as different introductions: i.e   \[(0=_{\mathbb{N}} 1)\longleftrightarrow \bot\]
• It would make sense to write \(\neg (0=_{\mathbb{N}}1)\) or more often \(0\neq 1\).

Many empty types;

One empty set

• If a mere proposition is empty, we have no evidence, but we still learn something about the claim, it is false.
• When a set is empty we have no idea what set is about, there is only one empty set so the properties of this set are about all false, claims, not helpful.
• So empty types are not the same but "equivalent" by functions \[(0=_{\mathbb{N}} 1)\longleftrightarrow \bot\]
• Empty sets however are all the same, and so we lost all the information.

Caution with formulas

• No evidence for \(0=_{\mathbb{N}}1\) as different introductions;
• But there is evidence  for \(0+0=_{\mathbb{N}} 0\), what is the difference?

Proving negatives

 

Definition. "Not P"

\[\neg P :\equiv P\to \bot\]

 

So evidence for \(\neg P\) means a function \(f:P\to \bot\).  Function elimination now recovers "contradiction":

 

 

\[\frac{P,\neg P}{\bot}\equiv \frac{e:P, f:P\to \bot}{f(e):\bot}\]

 

 

(Since \(\bot\to P\) this means \(P\leftrightarrow \bot\).)

 

WHAT THIS DOES NOT SAY!

 

\[\vdash P\sqcup \neg P\]

 

That is, we have not assumed either P or not P. (Known as "Law of Excluded Middle (LEM)")

 

Instead we are saying if P and not P then everything.

 

Ours is a strictly weaker logic;

but you wont need stronger logic in algebra.

"3 is not even"

Prove this in a way that can be verified.

Steps

• Clarify the mere proposition
• Give what "not true" might mean.
• Prove it in the example

Let \(P\) be a mere proposition.

E.g. \(P(n):\equiv \) "n is even"

\[\frac{n\in\mathbb{N}}{P(n):type}(F)\]

 

\[\frac{k\in\mathbb{N}}{witness(k):P(k+k)}(I)\]

 

\[\frac{e:P(n)}{testimony(e)\in\mathbb{N}}(E)\]

 

\[\frac{k\in\mathbb{N}\quad e:P(k+k)}{refl:k=testimony(witness(k))}(C)\]

 

\(P(3):\equiv \) "3 is even"  is a data type, it exists.

\[\frac{n\in\mathbb{N}}{P(n):type}(F)\]

 

 

BUT ...would the introduction rule ever let us create data of this type?

\[\frac{k\in\mathbb{N}}{witness(k):P(k+k)}(I)\]

 

A proposition without proof is still a proposition...but you might question if it is true!

 

Make a function \(P(3)\to \bot\).

Assume \(e:P(3)\), so \(e=witness(k):P(k+k)\) and \(k+k=3\).

 

• Case \(k=0\): \(0+0=0=3\).  So \(P(3)\to (0=3)\leftrightarrow \bot\).
• Case \(k=S(j)\): \[S(2)=3=S(j)+S(j)=S(S(j)+j).\]  So \(2=S(j)+j=j+S(j)=S(j+j)\).  Hence \(1=j+j\)
  • Case \(j=0\): \(1=0+0=0\); so,\[P(3)\to (1=0)\leftrightarrow \bot\]
  • Case \(j=S(\ell)\): \(1=S(\ell)+S(\ell)=S(S(\ell)+\ell)\) which means \(0=S(\ell)+\ell\) which as above reduces to \(\bot\).