convert biconditional to pseudo-boolean (inequality / equation) constraints I am working with pseudo-Boolean and I want to convert the bi-conditional $(a \land b) \iff c$ to inequality or equation. 
my attempt was.
First, convert the bi-conditional to two implies
$$((a \land b) \to c) \land (c \to (a \land b))$$
Then $$(\lnot(a \land b) \lor c ) \land (\lnot c \lor (a \land b))$$
Then, the left side part become 
$$(\lnot a \lor\lnot b\lor c)$$
and the right hand side part become 
$$((\lnot c \lor a) \land (\lnot c \lor b))$$
but to convert the right hand side part to constraints effected negatively on the pseudo-Boolean results.
So, my question is, am I correct in these previous steps or not? 
 A: A pseudo Boolean constraint is a constraint of the form, where $w_i$ are the weights, $x_i$ the Boolean variables, $k$ some comparison value and $\#$ some comparison operator:
$$\sum_{i=1}^n w_i x_i\,\#\,k$$
Since disjunction $x_1 \lor .. \lor x_n$ corresponds to the pseudo Boolean constraint $1 x_1 + ... + 1 x_n \geq 1$ and taking the negation of a boolean variable $x$ amounts to computing $1-x$ we can translate your Boolean formula into multiple pseudo Boolean constraints.
Since in conjunctive normal form your Boolean formula $(a \land b) \iff c$ amounts to the Boolean formula $(\lnot a\lor\lnot b\lor c)\land(\lnot c\lor a)\land(\lnot c\lor b)$ we can translate each clause and get:
$$(1-a)+(1-b)+c\geq 1$$
$$(1-c)+a\geq 1$$
$$(1-c)+b\geq 1$$
Or braught in the weighted sum form:
$$-1 a + -1 b + 1 c\geq -1$$
$$1 a + -1 c\geq 0$$
$$1 b + -1 c\geq 0$$
I dont know whether the alternate form, disjunctive normal form would be useful at all.
A: 
So, my question is, am I correct in these previous steps or not?

You are correct. By definition $X\leftrightarrow Y\equiv (\lnot X\lor Y)\land(\lnot Y\lor X)$, so... $$(a\land b)\leftrightarrow c ~\equiv~ (\lnot a\lor\lnot b\lor c)\land(\lnot c\lor a)\land(\lnot c\lor b)$$

Using the alternate definition that $X\leftrightarrow Y \equiv (X\land Y)\lor(\lnot X\land\lnot Y)$, will give... $$(a\land b)\leftrightarrow c ~\equiv~ ( a\land b\land c)\lor(\lnot a\land\lnot c)\lor(\lnot b\land\lnot c)$$
