NAND with Equal Arguments
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Theorem
Let $\uparrow$ signify the NAND operation.
Then, for any proposition $p$:
- $p \uparrow p \dashv \vdash \neg p$
That is, the NAND of a proposition with itself corresponds to the negation operation.
Proof 1
By the tableau method of natural deduction:
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:
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 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables:
- $\begin{array}{|ccc||cc|} \hline
p & \uparrow & p & \neg & p \\ \hline \F & \T & \F & \T & \F \\ \T & \F & \T & \F & \T \\ \hline \end{array}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\blacksquare$
Also see
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(5)$
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$