NAND is Commutative/Proof 1

Theorem

$p \uparrow q \dashv \vdash q \uparrow p$


Proof

By the tableau method of natural deduction:

$p \uparrow q \vdash q \uparrow p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \uparrow q$ Premise (None)
2 1 $\neg \paren {p \land q}$ Sequent Introduction 1 Definition of Logical NAND
3 1 $\neg \paren {q \land p}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $q \uparrow p$ Sequent Introduction 3 Definition of Logical NAND

$\Box$


By the tableau method of natural deduction:

$q \uparrow p \vdash p \uparrow q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \uparrow p$ Premise (None)
2 1 $\neg \paren {q \land p}$ Sequent Introduction 1 Definition of Logical NAND
3 1 $\neg \paren {p \land q}$ Sequent Introduction 2 Conjunction is Commutative
4 1 $p \uparrow q$ Sequent Introduction 3 Definition of Logical NAND

$\blacksquare$

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