Sequent Calculus:

the handwriting of logic

James B. Wilson Professor of Mathematics

Conjunction

\[\frac{\Gamma \vdash Q,P}{\Gamma \vdash P\wedge Q}\]

Dilemma: \[\begin{array}{c} P \vdash R \\ P\vdash R\\ P\vee Q \\ \hline R \end{array}\]

Logic Requires

A context (stuff at the top of a program, the books you were supposed to have read by now, conventions...)
A grammar (guarantee what we put in language can be read the same way by everyone)
Rules.

Aristotle

Syllogism: a largely poetic scheme with 3 parts.

"All men are mortal."
"Socrates is a man."
"Therefore Socrates is mortal."

First "syntax" based model of logic.

There are 256 options, but only 24 "valid", history of mnemonics "Barbara, Barbari,..."

(Major premise)

(Minor premise)

(Conclusion)

Frege & Gentzen

Ignore the poetry, expand the premises.

\[\frac{Premises}{Conclusion}\]

Line is pronounced "entails"

One-line notation

\(Premises ~\vdash~Conclusion\)

Frege had \(|\) for "judgement" and \(-\) for content.
Gentzen brought them together as \(\vdash\) to "judge content".

We call it "Turnstyle", typed U+22A2 or &vdash or \vdash

And

Pick a language & grammar, i.e. the symbol(s) for and how data will placed with the symbol.

<Conjunction> ::= both <Term> and <Term>

<Conjunction> ::= (<Term> && <Term>)

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

"both Jack and Jill"

(i<=j) && (j<=100)

\[(x^2=0)\wedge(x\geq 0)\]

And

(L)anguage

2. Make introduction.

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

\(P,Q \vdash P\wedge Q\qquad (I_{\wedge})\).

\[\begin{array}{rl} \Gamma &\vdash P\\ \Gamma &\vdash Q\\ \hline \Gamma & \vdash P \wedge Q\end{array}\qquad(I_{\wedge})\]

I for "intro", symbol to remind use of operator

Better yet, make context explicit as \(\Gamma\) or \(\texttt{ctx}\)

And

(L)anguage

(I)ntroduction

\[\begin{array}{rl} \Gamma &\vdash P\wedge Q\\ \hline \Gamma & \vdash P\end{array}(E_{L\wedge})\qquad\begin{array}{rl} \Gamma &\vdash P\wedge Q\\ \hline \Gamma & \vdash Q\end{array}(E_{\wedge R})\]

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

3. Rule(s) to eliminate symbol.

\(P,Q \vdash P\wedge Q\qquad (I_{\wedge})\).

\(P\wedge Q \vdash P\qquad (E_{L\wedge})\)

\(P\wedge Q \vdash Q \qquad (E_{\wedge R})\)

Using logical And

(L)anguage

(I)ntroduction

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

(E)limination

\(P,Q \vdash P\wedge Q\qquad (I_{\wedge})\).

\(P\wedge Q \vdash P\qquad (E_{L\wedge})\)

\(P\wedge Q \vdash Q \qquad (E_{\wedge R})\)

L.I.E. is a pattern for many logical operators

And

(L)anguage

(I)ntroduction

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

(E)limination

\(P,Q \vdash P\wedge Q\qquad (I_{\wedge})\).

\(P\wedge Q \vdash P\qquad (E_{L\wedge})\)

\(P\wedge Q \vdash Q \qquad (E_{\wedge R})\)

Measurements are absolute values. = \(\Gamma \vdash P:=(\ell\geq 0)\)

Area is 100. = \(\Gamma \vdash (\ell^2=100)\)

\(\Gamma \vdash (\ell\geq 0) \wedge (\ell^2=100)\)

What part of our AND "LIE" are we using?

And

(L)anguage

(I)ntroduction

\[\langle conj\rangle ::= (\langle term\rangle \wedge \langle term\rangle)\]

(E)limination

\(P,Q \vdash P\wedge Q\qquad (I_{\wedge})\).

\(P\wedge Q \vdash P\qquad (E_{L\wedge})\)

\(P\wedge Q \vdash Q \qquad (E_{\wedge R})\)

Lakes are deep and wide.

Lakes are deep

What part of our AND "LIE" are we using?

Logical Or?

(L)anguage

(I)ntroduction

\[\langle conj\rangle ::= (\langle term\rangle \vee \langle term\rangle)\]

(E)limination

\(P \vdash P\vee Q\qquad (I_{L \vee})\)

\(Q \vdash P\vee Q\qquad (I_{\wedge R})\)

\(P\vee Q \vdash P?\) or is it \(P\vee Q \vdash Q?\)

What a dilemma!

Disjunctive Dilemma

\[\begin{array}{rl} \ \Gamma,P &\vdash R \\ \Gamma,Q&\vdash R\\ \Gamma &\vdash P\vee Q\\ \hline \Gamma & \vdash R\end{array}\]

<disjunction> ::= either <term> or <term>

<disjunction> ::= <term> || <term>

Using Logical Or?

(L)anguage

(I)ntroduction

\[\langle conj\rangle ::= (\langle term\rangle \vee \langle term\rangle)\]

(E)limination

\(P \vdash P\vee Q\qquad (I_{L \vee})\)

\(Q \vdash P\vee Q\qquad (I_{\wedge R})\)

Disjunctive Dilemma

\[\begin{array}{rl} \ \Gamma,P &\vdash R \\ \Gamma,Q&\vdash R\\ \Gamma &\vdash P\vee Q\\ \hline \Gamma & \vdash R\end{array}\]

Wet roads lead to car crashes,

inexperience leads to car crashes, and

today there is a wet road or inexperienced driver then ...

...we will have a car crash.

And vs. Or

And and Or are similar and in many ways "dual":

And: 1 introduction 2 eliminations.
Or: 2 introductions 1 elimination.

However, there are some hints of non-dual nature, e.g. we cannot use Or without a third term.

Logical true?

(L)anguage

(I)ntroduction

\[\langle boolean\rangle ::= \top ~|~ \bot \]

(E)limination

\(\vdash \top\)

\(\bot\vdash P\)

<boolean> ::= true | false

Law of Explosion:

One false premise can conclude anything.

Used in classical and intuitionisitic logic, avoid in paraconsistent logic.

You can always assume truth is valid, so \(P\vdash \top\).

Logical not?

(L)anguage

(I)ntroduction

\[\langle neg\rangle ::= \neg <term> \]

(E)limination

\[\frac{\Gamma, P\vdash \bot}{\Gamma \vdash \neg P}\]

\[\begin{array}{rl} \Gamma & \vdash \neg P\\ \Gamma & \vdash P\\ \hline \Gamma & \vdash \bot\end{array}\]

<neg> ::= not <term>

If P leads to falsity, it cannot be true.

<neg> ::= !<term>

\[\frac{ P\vdash \bot}{ \vdash \neg P}\]

\[\frac{\Gamma\vdash ~(\neg P) \wedge P}{\Gamma\vdash ~\bot}\]

Premise "P" leads to falsity \(\bot\)

Conclusion "not P" is valid.