Question

PDNF of p->((p->q)^~(~qv~p))

Answer 1

Transform the expression:

$P \to ((P \to Q) \land \lnot(\lnot Q \lor \lnot P))$

Expressing the conditional using logical-or and using De Morgan's law:

$P \to ((\lnot P \lor Q) \land Q \land P)$

Using the absorbtion law:

$P \to (Q \land P)$

Expressing the conditional using the logical or:

$\lnot P \lor (Q \land P)$

Using the distibutive law:

$(\lnot P \lor Q) \land (\lnot P \lor P)$

$(\lnot P \lor Q) \land T$

$\lnot P \lor Q$

Make a truth table:

$P \ \ \ \ Q \ \ \ \ \ \lnot P \lor Q$

$F \ \ \ \ F \ \ \ \ \ T$

$F \ \ \ \ T \ \ \ \ \ T$

$T \ \ \ \ F \ \ \ \ \ F$

$T \ \ \ \ T \ \ \ \ \ T$

For the rows that have the expression true do the following to obtain the PDNF:

For each row right down all the variables in the expression, if the variable has False value in this row, negate it. Make a conjunction of these terms.

Each such conjuction unite in the complete expression using a disjunction.

For the required expression there are 3 such conjunctions:

$(P∧Q)$

$(¬P∧Q)$

$(¬P∧¬Q)$

Therefore, the PDNF is:

$(P∧Q)∨(¬P∧Q)∨(¬P∧¬Q)$

Answer: $(P∧Q)∨(¬P∧Q)∨(¬P∧¬Q)$