Is this a valid proof of $(A∧B’) ∧C↔(A∧C) ∧B’$? So I am supposed to prove $(A∧B’) ∧C↔(A∧C) ∧B’$ using wffs and equivalence rules. I have never done such proof, and I want to check if my steps are correct. This assignment is only graded based off of completion, but I want to be sure I am understanding the concepts correctly. Thanks for all the help in advance. 
 Prove
       (A∧B’) ∧C↔(A∧C) ∧B’
       (A∧B’) ∧C = P
       (A∧C) ∧B’ = Q

       1. (A∧B’)          hyp
       2. C               hyp
       3. (A∧C)           hyp
       4. B’              hyp
       5. A ,B’,C         1,2 sim
       6. P,Q             5,  sim
       7.  P∧S            6, con
       8.  S∧P            7, comm
       9. (P→Q)∧(Q→P)     7,8 equ
       10. P↔Q            9 equ         

 A: You don't define $S$ anywhere, so I don't know how it ends up in 7 & 8!
But back up to the start. 
$$\begin{align} (A \land B') \land C &\iff A\land (B' \land C) \tag{associativity}\\ \\
& \iff A \land (C \land B') \tag{commutativity}\\ \\
&\iff (A\land C) \land B' \tag{associativity}
\end{align}$$
A: I learned it this way.
For this:
$$(A \land B') \land C \leftrightarrow (A\land C)\land B' $$
You need to show \begin{align}(A \land B') \land C \rightarrow (A\land C)\land B'\tag{1} \end{align} and \begin{align}(A\land C)\land B'\rightarrow (A \land B') \land C\tag{2}\end{align}
First show (1):
Use conditional proof: 


*

*Assume $(A\land B')\land C$

*With Simp1 and 1. you get $(A\land B')$

*With Simp1 and 2. you get $A$

*With Simp2 and 2. you get $B'$

*With Simp2 and 1. you get $C$

*With Kon, 3. and 5. you get $A\land C$

*With Kon, 4. and 6 you get $(A\land C)\land B'$

*With Conclusion of 1-7 you get $(A \land B') \land C \rightarrow (A\land C)\land B'$ 


Thus (1)
(2) can be proofen analogy
With (1) and (2) you get your equivalence!
