1. ## disjunctive normal form

I know that the disjunctive normal form is not unique.

<Start of excerpt from Wolfram MathWorld>
A statement is in disjunctive normal form if it is a disjunction (sequence of ORs) consisting of one or more disjuncts, each of which is a conjunction (AND) of one or more literals (i.e., statement letters and negations of statement letters; Mendelson 1997, p. 30). Disjunctive normal form is not unique.
<End of excerpt from Wolfram MathWorld>

For my purposes, I repeat, for emphasis, one sentence from the above excerpt:
Disjunctive normal form is not unique

OK.

( I use "~" to denote Logical-Negation, "+" to denote Logical-OR (inclusive), and "*" to denote Logical-AND )

I am _not_ looking for a formal proof of what I trying to do below; informality is sought.

I have a statement in conjunctive normal form, and I can express that statement in disjunctive normal form.

(~P + Q) * (P + ~Q) (Conjunctive Normal Form)
Using the Law of Distribution, this statement can be converted to
(~P * P) + (~P * ~Q) + (Q * P) + (Q * ~Q)
Because of the Law of Contradiction, this statement can be converted to
FALSE + (~P * ~Q) + (Q * P) + FALSE
which can be converted to
(~P * ~Q) + (Q * P) (which is in disjunctive normal form)

I want to arrive at another, logically equivalent, statement in disjunctive normal form :
(P * ~Q) + (~P * Q)

How do I get, informally,
From: (~P * ~Q) + (Q * P)
To: (P * ~Q) + (~P * Q)

2. Originally Posted by brooks972
I know that the disjunctive normal form is not unique.

<Start of excerpt from Wolfram MathWorld>
A statement is in disjunctive normal form if it is a disjunction (sequence of ORs) consisting of one or more disjuncts, each of which is a conjunction (AND) of one or more literals (i.e., statement letters and negations of statement letters; Mendelson 1997, p. 30). Disjunctive normal form is not unique.
<End of excerpt from Wolfram MathWorld>

For my purposes, I repeat, for emphasis, one sentence from the above excerpt:
Disjunctive normal form is not unique

OK.

( I use "~" to denote Logical-Negation, "+" to denote Logical-OR (inclusive), and "*" to denote Logical-AND )

I am _not_ looking for a formal proof of what I trying to do below; informality is sought.

I have a statement in conjunctive normal form, and I can express that statement in disjunctive normal form.

(~P + Q) * (P + ~Q) (Conjunctive Normal Form)
Using the Law of Distribution, this statement can be converted to
(~P * P) + (~P * ~Q) + (Q * P) + (Q * ~Q)
Because of the Law of Contradiction, this statement can be converted to
FALSE + (~P * ~Q) + (Q * P) + FALSE
which can be converted to
(~P * ~Q) + (Q * P) (which is in disjunctive normal form)

I want to arrive at another, logically equivalent, statement in disjunctive normal form :
(P * ~Q) + (~P * Q)

How do I get, informally,
From: (~P * ~Q) + (Q * P)
To: (P * ~Q) + (~P * Q)
Well, you can't: because $\displaystyle (\neg p \wedge \neg q)\vee (p\wedge q)$ does not imply $\displaystyle (p\wedge \neg q)\vee (\neg p\wedge q)$.
Proof: Set both p and q true. In that case $\displaystyle (\neg p \wedge \neg q)\vee (p\wedge q)$ is true, but $\displaystyle (p\wedge \neg q)\vee (\neg p\wedge q)$ is false.