NOR is Commutative

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Theorem

Let $\downarrow$ signify the NOR operation.


Then, for any two propositions $p$ and $q$:

$p \downarrow q \dashv \vdash q \downarrow p$

That is, NOR is commutative.


Proof 1

By the tableau method of natural deduction:

$p \downarrow q \vdash q \downarrow p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \downarrow q$ Premise (None)
2 1 $\neg \paren {p \lor q}$ Sequent Introduction 1 Definition of Logical NOR
3 1 $\neg \paren {q \lor p}$ Sequent Introduction 2 Disjunction is Commutative
4 1 $q \uparrow p$ Sequent Introduction 3 Definition of Logical NOR

$\Box$


By the tableau method of natural deduction:

$q \downarrow p \vdash p \uparrow q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \downarrow p$ Premise (None)
2 1 $\neg \paren {q \lor p}$ Sequent Introduction 1 Definition of Logical NOR
3 1 $\neg \paren {p \lor q}$ Sequent Introduction 2 Disjunction is Commutative
4 1 $p \downarrow q$ Sequent Introduction 3 Definition of Logical NOR

$\blacksquare$


Proof by Truth Table

Apply the Method of Truth Tables:

$\begin{array}{|ccc||ccc|} \hline p & \downarrow & q & q & \downarrow & p \\ \hline \F & \T & \F & \F & \T & \F \\ \F & \F & \T & \T & \F & \F \\ \T & \F & \F & \F & \F & \T \\ \T & \F & \T & \T & \F & \T \\ \hline \end{array}$


As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\blacksquare$


Also see


Sources