Conjunction in terms of NAND
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Theorem
- $p \land q \dashv \vdash \paren {p \uparrow q} \uparrow \paren {p \uparrow q}$
where $\land$ denotes logical conjunction and $\uparrow$ denotes logical NAND.
Proof
| \(\ds p \land q\) | \(\dashv \vdash\) | \(\ds \neg \neg \paren {p \land q}\) | Double Negation | |||||||||||
| \(\ds \) | \(\dashv \vdash\) | \(\ds \neg \paren {p \uparrow q}\) | Definition of Logical NAND | |||||||||||
| \(\ds \) | \(\dashv \vdash\) | \(\ds \paren {p \uparrow q} \uparrow \paren {p \uparrow q}\) | NAND with Equal Arguments |
$\blacksquare$
Sources
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 2.5$: Further Logical Constants
- 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): $\S 2.4.2$