Boolean Equation Equivalence Problem

[jomanscool2]

Homework Statement

Part a was to prove the equivalence of the two equations using a truth table. Done.
Part b is to prove the equation using the 10 properties of boolean logic, as seen in http://en.wikipedia.org/wiki/Boolean_logic#Properties"

We are trying to prove:
(p∨q)∧(¬q∨r) ⇔ (p∧r)∨((p∧(¬q∧¬r))∨(¬p∧(q∧r))).

Homework Equations

The Attempt at a Solution

I have done one solution, which was basically a brute force attempt at the problem including lots of distribution over and over again. It was over two pages long and about 40 steps. Not ideal, and not full credit for the problem either. My current attempt I am just in need of a way to 'legally' add in a "not p" into the second term of the first line in the second block of expressions. The first block is working from the longer equation to the short one, and the other block is from the shorter to the longer, trying to meet in the middle.

I have no idea if this will lead to a solution, but I am being hopeful!

(p∧r)∨((p∧(¬q∧¬r))∨(¬p∧(q∧r)))
(p∧(r∨(¬q∧¬r)))∨(¬p∧(q∧r))
(p∧((r∨¬r)∧(r∨¬q)))∨(¬p∧(q∧r))
(p∧(T_0∧(r∨¬q)))∨(¬p∧(q∧r))
(p∧(r∨¬q))∨(¬p∧(q∧r ))

(p∧(r∨¬q))∨(q∧r)
(p∧(r∨¬q))∨(F_0∨(q∧r))
(p∧(r∨¬q))∨((q∧¬q)∨(q∧r))
(p∧(r∨¬q))∨(q∧(r∨¬q))
(p∨q)∧(r∨¬q)

I am just wondering if anyone has any suggestions on how to get that last step on the proof I just posted, or any guidance of another short (half a page typed) proofs for this equation.

Edit: the half that contains the difference is not equal on a truth table, meaning it cannot be directly converted. If I were to connect those two equations, I believe that the entire equation would have to be manipulated.

 

[story645]

Start by posting which properties you're using in each stage. You'll quickly see that you're only using about 4 and dropping out two key properties that'll go a long way towards getting you to a solution.