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 | falseLaw 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!
Logic of OR TRUE AND NOT
By James Wilson
Logic of OR TRUE AND NOT
We are making our own logic so how hard can it be to make Or, True and Not? Surprises exist because we wont simply be shopping for these concepts already buried somewhere else. We are making them all natural from scratch! While I love these constructs I find them both subtle and worthy of multiple perspectives.
