Not true
2020 James B. Wilson
Colorado State University
Objective
Proving something is not true
Not an Objective
Proving something is true
(This will make more sense at the end.)
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.
Programmers know this type as "void", i.e.
void f(int i){...}
returns nothing because there is no data of type void
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?
Difference: \(0+0\) is a formula and it must be computed (\(\beta\)-reduced) before comparing. Since \[0+0\rhd 0\] by our definition of + we do indeed end with \(0=_{\mathbb{N}} 0\).
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 \(\bot\).
\(\sqrt{5}\) is irrational.
- \[\neg(\sqrt{5}\in \mathbb{Q})\equiv \bigsqcup_{a\in \mathbb{Z}}\bigsqcup_{b\in \mathbb{Z}^+}(\sqrt{5}=a/b)\to \bot\]
- Given evidence for \(\sqrt{5}=a/b\) with \(a/b\) reduced (property \(Q(a/b)\)); then there is evidence for \(5b^2=a^2\) so \(5\mid a^2\);
- hence \(5\mid a\) and \(25\mid a^2\);
- hence \(5\mid b\) so \(a/b\) not reduced \(\neq Q(a/b)\).
- \[\frac{Q(a/b),\neg Q(a/b)}{\bot}\]
There are infinitely many primes
- \(\neg P(n)\) where \(P(n)\equiv\) "q prime implies \(q\in \{p_1,\ldots,p_n\}\)"
- Assume \(e:P(n)\)
- Make \(m=p_1\cdots p_n+1\).
- Proved already that \(gcd(n,n+1)=1\)
- So no prime \(p_i\) divides \(m\)
- Any prime q dividing \(m\) is not in \(\{p_1,\ldots,p_n\}\), i.e. \(\neg P(n)\).
- \[\frac{P(n),\neg P(n)}{\bot}\]
Terminology:
-
Proof of negation: Want to prove \(\neg P\equiv P\to \bot\)
- Assume hypothesis, i.e. \(P\), derive conclusion, only here the conclusion \(\bot\) has no evidence so deriving conclusion means to steer towards a type known to be empty.
-
Proof by contradiction: Want to prove \(P\).
- Assume \(\neg P\), work towards \(Q,\neg Q\).
These are different. The second requires an axiom of logic (LEM). So proofs of negation hold in more places than proofs by contradiction.
Godel's Negative Embedding Theorem
Intuitionistic logic cannot disprove anything in classical logic.
"...its not (not true) that..." still weaker than "...it is true..." but close enough for us.
Intuitionistic Logic: no contradiction
Classical Logic: with contradiction.
"Proof by contradiction is dangerous, because if you make a mistake it helps you." -J.P. Serre
Not true
By James Wilson
Not true
proving negatives
