Construct the disjunctive normal form of the proposition (p → q) ∧ (r ↔ p).

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We have to find the DNF of (p → q) ∧ (r ↔ p)

We know the basic equivalences

i.e. p → q is equivalent to ~p ∨ q

and r ↔ p is equivalent to (r → p) ∧ (p → r)

Hence DNF of above proposition can be resolved as,

(~p ∨ q) ∧ ((r → p) ∧ (p → r))

= (~p ∨ q) ∧ (~r ∨ p) ∧ (~p ∨ r)

Truth table of above expression can be

From this table we can easily get the DNF by selecting only the True terms.

Hence DNF is:

(p ∧q ∧ r) ∨ (p ∧ q∧ ~ r) ∨ (~ p ∧ q ∧ ~r) ∨ (~p∧~q∧~r)