0
$\begingroup$

a) Find a representation of each of the following statements using only the logical connective nand.
i. $\neg p$
ii. $p \wedge q$
iii. $p \vee q$
iv. $p \rightarrow q$
v. $p \leftrightarrow q$

b) Explain why every statement has a representation using only the logical connective nand.

My work so far:

i. $\neg p \Leftrightarrow p \barwedge p$
iii. $p \vee q \Leftrightarrow (p \barwedge p) \barwedge (q \barwedge q)$

$\endgroup$
4
  • $\begingroup$ Yes. So far, so good. What problem do you have with the otters? $\endgroup$ – Graham Kemp Sep 21 '17 at 6:45
  • $\begingroup$ My problem is I can't figure out what they are $\endgroup$ – Shea Sep 21 '17 at 6:45
  • $\begingroup$ Why not? You managed those two just fine. What have you tried with the others? $\endgroup$ – Graham Kemp Sep 21 '17 at 6:48
  • $\begingroup$ I don't know where to start with the others, that's the problem. $\endgroup$ – Shea Sep 21 '17 at 6:56
0
$\begingroup$

$\def\nand{\barwedge}$

i. $\neg p$

Since $p=p\wedge p$ , then indeed $\neg p = p\nand p$ $\checkmark$

ii. $p \wedge q$

Well, $p\wedge q = \neg(p\nand q)$ so...

iii. $p \vee q$

Yes, $p\vee q = \neg(\neg p\wedge \neg q) = \neg p\nand \neg q=(p\nand p)\nand(q\nand q)$ $\checkmark$

iv. $p \rightarrow q$

Well $p\to q = \neg p\vee q$ so...

v. $p \leftrightarrow q$

Likewise $p\leftrightarrow q = (p\to q)\wedge (q\to p)$ so...

$\endgroup$
8
  • $\begingroup$ We are only allowed to use $\barwedge$, no other operators can be used. I know how to solve all five of them if I am allowed to use $\neg$, but I can't. $\endgroup$ – Shea Sep 21 '17 at 6:58
  • $\begingroup$ But you know that $\neg f(p,q)=f(p,q)\nand f(p,q)$... It's the first one you did, and you used that very fact to express $p\vee q$ as $(p\nand p)\nand (q\nand q)$ $\endgroup$ – Graham Kemp Sep 21 '17 at 7:00
  • $\begingroup$ @Shea But You've already showed haw to replace negation with nand. $\endgroup$ – Jaroslaw Matlak Sep 21 '17 at 7:04
  • 1
    $\begingroup$ ...just substitute... $\neg x=x\nand x$ when $x$ is $(p\land q)$ $\endgroup$ – Graham Kemp Sep 21 '17 at 7:05
  • 2
    $\begingroup$ @Shea and if we take another statement $a\equiv p\wedge q$, then how do you write $\neg a$ using only nand? $\endgroup$ – Jaroslaw Matlak Sep 21 '17 at 7:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.