Using induction to prove an iff statement?

I am working on this problem on Enderton's book:

1.2.11. Show that a truth assignment $v$ satisfies the wff (...($A_1$ $\leftrightarrow$ $A_2$)$\leftrightarrow$ ... $\leftrightarrow$ $A_n$) iff $v(A_i)$ = $F$ for an even number of $i$'s, 1 $\leq$ $i$ $\leq$ $n$. (By the associative law of $\leftrightarrow$, the placement of the parentheses is not crucial.)

I intend to approach this problem using proof by induction. Let $k$ denote the number of $i$'s. Then we use induction on $k$ for $k$ = 0, 2, 4, 6, ... up until $n$ or $n - 1$ depending on the parity of $n$. Here is my problem: Does this approach satisfy the requirement of the exercise, namely the iff condition? Or does my approach only just prove one direction of the exercise?

Follow-up question in case the answer for the above question is No: How do I go from having a truth assignment $v$ that satisfies the wff to showing that $k$ is even? Is using proof by contradiction a good idea? For example, assume that $v$ is a truth assignment that satisfies the wff, but $k$ is odd. Then I somehow show the contradiction?

Solution 1:

Your induction on $k$ which is only even is not valid as $k$ may very well likely to be odd such as the case $(A_1 \leftrightarrow A_2)\leftrightarrow A_3$, instead you can just simply do induction on $n$, and let $k$ be the number of $i$'s where $v(A_i)=F$. Base case when $n=1$ is trivial, as $A_1 \iff k=0$. For the general inductive case assuming $(A_1 \leftrightarrow A_2)\leftrightarrow ...A_n \iff k=even$, using proof by contradiction suppose $k$ is odd for the $(n+1)$ step, then clearly $v(A_{n+1})=F$, but immediately $((A_1 \leftrightarrow A_2)\leftrightarrow ...A_n)\nleftrightarrow A_{n+1}$ (we also have the placement of the parentheses is not crucial as hinted) which contradicts the expectation. So following this line and you're right that we don't need prove back-and-forth two directions separately.