exam question

Hi, in the past paper we were given , question 1b says:
Prove that the Boolean operators negation, implication and disjunction can also be defined from (up arrow), arguing directly from the inductive definition of truth values of propositional formulas.

Anyone have any idea how to answer this?

Here is an example of a similar question and answer:

Question: Prove that disjunction and implication can be defined in terms of conjunction and negation.

To prove this, you first give a definition of disjunction which uses only conjunction and negation:

P \/ Q is equivalent to ~(~P /\ ~Q)

You then prove that this equivalence holds:

P \/ Q is true iff P is true or Q is true
iff ~P is false or ~Q is false
iff ~P /\ ~Q is false
iff ~(~P /\ ~Q) is true

Now you can define implication using disjunction (which you have just shown can be defined in terms of conjunction and negation) and negation:

P -> Q is equivalent to ~P \/ Q

You then prove that this equivalence holds, similar to the proof above, and you're done.

For Question 1b, you start off only with NAND. Question 1a shows that AND can be defined in terms of NAND. For NOT, see the equivalence given on page 21. So now you've got NAND (which you started off with), AND (which you defined in 1a), and NOT. Next you define disjunction in terms of these operators, and prove that your equivalence holds. Then you define implication, and you're done.

Thank you