OR TRUE NOT in Sequent Calculus

James B. Wilson Professor of Mathematics

Conjunction

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

 

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

Sequent Calculus separate premises from conclusion

  • A horizontal line or turnstyle \(\vdash\) is the separator, called "entails"
  • Logical operators occurs when we specify a 
    • (L)anguage & Grammar
    • (I)ntroduction rules (how to introduce the operator)
    • (E)limination rules (how to use/eliminate the operator)

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

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.

Logical implication?

(L)anguage

  

(I)ntroduction

\[\langle implies\rangle ::= <term>\Rightarrow <term>\]

(E)limination

\[\begin{array}{rl} \Gamma & \vdash P\Rightarrow Q\\ \Gamma & \vdash P\\ \hline \Gamma & \vdash Q\end{array}\]

<implies> ::= if <term> then <term>

 

\[\frac{ \Gamma, \overbrace{P,\ldots,P}\vdash Q}{\Gamma \vdash P\Rightarrow Q}\]

 

"P" appears any number of times, even 0 times!