NAND with Equal Arguments/Proof 1

Theorem

$p \uparrow p \dashv \vdash \neg p$

That is, the NAND of a proposition with itself corresponds to the negation operation.


Proof

By the tableau method of natural deduction:

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

$\Box$


By the tableau method of natural deduction:

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

$\blacksquare$

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