Disjunction in terms of NAND
Jump to navigation
Jump to search
Theorem
- $p \lor q \dashv \vdash \paren {p \uparrow p} \uparrow \paren {q \uparrow q}$
where $\lor$ denotes logical disjunction and $\uparrow$ denotes logical NAND.
Proof
\(\ds p \lor q\) | \(\dashv \vdash\) | \(\ds \neg \paren {\neg p \land \neg q}\) | De Morgan's Laws (Logic): Disjunction | |||||||||||
\(\ds \) | \(\dashv \vdash\) | \(\ds \neg p \uparrow \neg q\) | Definition of Logical NAND | |||||||||||
\(\ds \) | \(\dashv \vdash\) | \(\ds \paren {p \uparrow p} \uparrow \paren {q \uparrow q}\) | 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 $(5)$