Conditional in terms of NAND/Proof 1
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Theorem
- $p \implies q \dashv \vdash p \uparrow \paren {q \uparrow q}$
Proof
\(\ds p \implies q\) | \(\dashv \vdash\) | \(\ds \neg \paren {p \land \neg q}\) | Conditional is Equivalent to Negation of Conjunction with Negative | |||||||||||
\(\ds \) | \(\dashv \vdash\) | \(\ds p \uparrow \neg q\) | Definition of Logical NAND | |||||||||||
\(\ds \) | \(\dashv \vdash\) | \(\ds p \uparrow \paren {q \uparrow q}\) | NAND with Equal Arguments |
$\blacksquare$