I was able to find the answer to (a) and (b), but I\'d love some help on (c) and

Question

I was able to find the answer to (a) and (b), but I'd love some help on (c) and (d). The answers to (a) and (b) may be helpful for (c) and (d). Thanks.

3. All You Need is Nand, Nand, Nand... (20 points) The NAND connective takes two propositions and evaluates to False when both propositions are True and evaluates to True otherwise. In circuit diagrams, the gate for NAND is denoted by NAN HARI The NAND of p and q is written as p |q. Demonstrate that we can construct all the other connectives by just using NAND by writing propositional formulae for each of the following while only using NAND connectives and no constants like T or F Hint: It's okay to use a single input/output more than once. (a) [5 Points] (b) [5 Pointsl p Vq (c) [5 Points) pAg (d) [5 Pointsl pq

Explanation / Answer

Assumption: | denotes NAND.

a) ¬p = p | p

b) p V q = (p | p) | (q | q)

c) p / q = ¬(p | q) = (p | q) | (p | q)

d) p<-> q = [ ( p NAND p ) NAND ( q NAND q ) ] NAND( p NAND q )

= [ ( p | p ) | ( q | q ) ] | ( p | q )

Explanation:

C) p AND q = NOT (p NAND q)

Now NOT A is A | A we know from (a) so NOT (p NAND q) is = (p NAND q) NAND (p NAND q) which is = (p | q) | (p | q)

D) p<-> q is logically equivalent to p X-NOR q and if we can represent p X-NOR q using NAND then we can get answer.

Assumption: ‘ represents NOT.

p X-NOR q is = (p' q ' + pq) = ((p' q')' (p q)')'

= [ ( p NAND p ) NAND ( q NAND q ) ] NAND( p NAND q )

Hence the answer is: p<-> q = [ ( p | p ) | ( q | q ) ] | ( p | q )

/*Hope this helps. Thanks.*/

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