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\title {Logic I \\ Lecture 18}

\maketitle

# Lecture 18

\def \ititle {Logic I}
\def \isubtitle {Lecture 18}
\begin{center}
{\Large
\textbf{\ititle}: \isubtitle
}

\iemail %
\end{center}
Readings refer to sections of the course textbook, \emph{Language, Proof and Logic}.

let’s get meta

\begin{minipage}{\columnwidth}
\textbf{If A ⊢ B then ⊢ A→B}
Proof Given a proof for A ⊢ B …
… we can turn it into a proof for ⊢ A→B:
\end{minipage}
\textbf{If ⊢ A→B then A ⊢ B}
\begin{minipage}{\columnwidth}
\textbf{If A ⊢ B then A ⊢ ¬¬B}
Proof:
\end{minipage}
\textbf{If A ⊢ C then A ⊢ B→C}
\textbf{If A ⊢ B and A ⊢ ¬C then A ⊢ ¬(B→C)}

## A ⊨ ⊥

\section{A ⊨ ⊥}

\section{A ⊨ ⊥}
We know what A ⊢ ⊥ means: it means that there's a proof from premises A of ⊥.

A ⊢ ⊥

What does putting the line through mean? Just that there's no proof of ⊥ from premises A.

A ⊬ ⊥

This is also easy to understand ...

A ⊨ B

A ⊭ B

An argument is logically valid just if there’s no possible situation in which the premises are true and the conclusion false.

But what does A ⊨ ⊥ mean?
The argument is logically valid ... no possible situation where ... but ⊥ is always false, so no possible situation where premises A are all true

A ⊨ ⊥

A ⊭ ⊥

ATT

ATT

## If (A ⊬ ⊥ entails A ⊭TT ⊥) then (A ⊨TT B entails A ⊢ B)

\section{If (A ⊬ ⊥ entails A ⊭TT ⊥) then (A ⊨TT B entails A ⊢ B) }

\section{If (A ⊬ ⊥ entails A ⊭TT ⊥) then (A ⊨TT B entails A ⊢ B) }
First we show that if \textbf{A}, ¬B ⊢ ⊥ then \textbf{A} ⊢ B .
(To do this, we show how to turn transform a proof of the former kind into a proof of the latter kind.)
Now assume that for all sets \textbf{A}, \textbf{A} ⊬ ⊥ entails \textbf{A} ⊭$_{TT}$ ⊥
Suppose that \textbf{A} ⊨TT B
Then \textbf{A}, ¬B ⊨TT ⊥
So, by our assumption, \textbf{A}, ¬B ⊢ ⊥
Therefore (by the thing we showed first) \textbf{A} ⊢ B

IfA, ¬B ⊢ ⊥ then A ⊢ B

 ​ 1 A1 ​ 2 ... ​ 3 An ​ 4 ¬B ​ 5 ... ​ 6 ⊥
1.

A1

2.

...

3.

An

 ​ 4 ¬B ​ 5 ... ​ 6 ⊥
7.¬¬B¬Intro: 4-6
8.B¬Elim: 7

Assume that for all sets A, A ⊬ ⊥ entails ATT

Suppose that A TT B

(We want to show that A ⊢ B)

Therefore: A, ¬B ⊨ TT

An argument is logically valid just if there’s no possible situation in which the premises are true and the conclusion false.

Suppose for a contradiction that A, ¬B ⊬ ⊥

Then, from the assumption, we would have A, ¬B ⊭ TT

So: A, ¬B ⊢ ⊥

Therefore (by the previous result) A ⊢ B

So we assumed this.
And derived this.

So:

If A ⊬ ⊥ entails ATT

then A TT B entails A ⊢ B

## The Essence of the Completeness Theorem

\section{The Essence of the Completeness Theorem}

\section{The Essence of the Completeness Theorem}

## Lemma for the Completeness Theorem

\section{Lemma for the Completeness Theorem}

\section{Lemma for the Completeness Theorem}
If for every sentence letter, P, either A ⊢ P or A ⊢ ¬P, then for every formula, X, either A ⊢ X or A ⊢ ¬X.
Proof
\textbf{Step a.} Suppose (for a contradiction) that there are formulae, X, such that A ⊬ X and A ⊬ ¬X. Take a shortest such formula, call it Y.
\textbf{Step b.} This formula, Y, must have one of the following forms: ¬P, P∨Q, P∧Q, P→Q, P↔Q, ⊥
\textbf{Step c.} We can show that whichever form X has, either A ⊢ Y and A ⊢ ¬Y.
Case 1: X is P→Q. Then since P and Q are shorter than X, either:
\hspace{5mm} (i) A ⊢ P and A ⊢ ¬Q
\hspace{5mm} or
\hspace{5mm} (ii) A ⊢ ¬P
\hspace{5mm} or
\hspace{5mm} (iii) A ⊢ Q
\hspace{5mm} If (i), A ⊢ ¬(P→Q), that is, A ⊢ ¬X.
\hspace{5mm} If (ii), A ⊢ P→Q, that is, A ⊢ ¬X.
\hspace{5mm} If (iii), A ⊢ P→Q, that is, A ⊢ ¬X.
\hspace{5mm} (Here we use the last two Proofs about Proofs, see earlier)
Case 2: X is ¬P.
\hspace{5mm} Then since P is shorter, A ⊢ P or A ⊢ ¬P.
\hspace{5mm} If A ⊢ P then A ⊢ ¬¬P so A ⊢ ¬X which would contradict our assumption. This is shown in the proofs about proofs above.
\hspace{5mm} If A ⊢ ¬P then A ⊢ X (because X is ¬P), which would contradict our assumption.
Case 3: …
\textbf{Step d.} The demonstration in Step c contradicts our assumption, so we can conclude that it is false. That is, either A ⊢ X and A ⊢ ¬X for every formula X.

## Proof of the Completeness Theorem

\section{Proof of the Completeness Theorem}

\section{Proof of the Completeness Theorem}

## Proof of Proposition 4 for the Completeness Theorem

\section{Proof of Proposition 4 for the Completeness Theorem}

\section{Proof of Proposition 4 for the Completeness Theorem}
This is a proof of Proposition 4 in §17.2 of the textbook LPL.

Define a structure, h, so that:

h(P)=T when P is in A*

h(P)=F when ¬P is in A*

Claim: h(X)=T for every X such that A* ⊢ X

We want to provide this slightly stronger claim (because we're going to do induction)

Proposition: for every X, h(X)=T iff A* ⊢ X

Suppose not. Then take the shortest sentence, Y, such that the Proposition is false.

Either: h(Y)=T and A* ⊬ Y

Or: h(Y)=F and A* ⊢ Y

We know that Y cannot be a sentence letter or its negation from the assumption about how h() was defined

Also, Y has the form (A→B) or (A∧B) or ...

Let's consider a case.

h(Y)=T

So h(A∧B)=T

So h(A)=T and h(B)=T

But A and B are shorter than Y

So, by assumption, A* ⊢ A and A* ⊢ B

Then A* ⊢ A∧B

i.e A* ⊢ Y

A* ⊢ Y i.e. A* ⊢ A→B

h(Y)=F i.e. h(A→B)=F

So h(A)=T and h(B)=F

But A and B are shorter than Y.

So, by assumption, A* ⊢ A and A* ⊬ B

But for all X, A* ⊢ X or A* ⊢ ¬X

So A* ⊢ ¬B

So A* ⊢ ¬(A→B) (proofs about proofs)