Conditional in terms of NAND/Proof 2
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Theorem
- $p \implies q \dashv \vdash p \uparrow \paren {q \uparrow q}$
Proof
| \(\ds p \implies q\) | \(\dashv \vdash\) | \(\ds \neg p \lor q\) | Rule of Material Implication | |||||||||||
| \(\ds \) | \(\dashv \vdash\) | \(\ds \neg p \lor \neg \neg q\) | Double Negation Introduction | |||||||||||
| \(\ds \) | \(\dashv \vdash\) | \(\ds p \uparrow \neg q\) | NAND as Disjunction of Negations | |||||||||||
| \(\ds \) | \(\dashv \vdash\) | \(\ds p \uparrow \left({q \uparrow q}\right)\) | 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