Equivalent properties of a proper ideal of a generalized boolean algebra I do not understand the item c) of the following question, the exercise 9 from section 1.2 from the book "Lattice-ordered Rings and Modules" from Stuart A. Steinberg:
A generalized boolean algebra is a distributive lattice $X$ with least element $0$ which is relatively complemented; this means that complements exist in each closed interval $[a,b]=\{x\in X:a\leq x\leq b\}$.
An ideal of a lattice $X$ is a nonempty subset $I$ such that if $a,b\in I$ and $c\in X$ with $c\leq a$, then $a\sqcup b\in I$ and $c\in I$. $I$ is a prime ideal of $X$ if and only if $I$ is proper ideal and whenever $a,b\in X$ with $a\sqcap y\in I$, then $a\in I$ or $b\in I$. Let $I$ be a subset of the generalized boolean algebra $X$. Show that $I$ is an ideal of $X$ if and only if $I$ is an ideal of the [induced boolean] ring $X$. Moreover, the following are equivalent for the proper ideal $I$:
a) $I$ is a prime ideal.
b) $I$ is a maximal ideal.
c) If $x\in X$ and $y$ is its complement in some interval $[a,b]$, then $x\in I$ or $y\in I$.
d) $|X/I|=2$.
Just help me to interpret correctly the item (c) (I was able to prove the equivalence of the other items).
 A: I think there might be a typo on the book. Indeed, condition $c)$ is not equivalent to the others. It implies that $I=X$. Indeed, we can take $x=y=a=b$ and so $a \in I$ for all $a \in X$. Thus there is no proper ideal satisfying $c)$. Whereas there are clearly proper ideals satisfying $a),b)$ and $d)$.
I think he meant the following condition:
$c')$ If $x \in X$ and $y$ is its complement in some interval $[0,b]$, then $x \in I$ or $y \in I$. 
Indeed, we can prove it is equivalent to $a)$:
$a) \Rightarrow c').$ If $x \in X$ and $y$ is its complement in some interval $[0,b]$, then $x \wedge y=0 \in I$. Since $I$ is prime, $x \in I$ or $y \in I$.
$c') \Rightarrow a).$ Let $x,y \in X$ such that $x \wedge y \in I$. Let $z$ be the complement of $y$ in $[0, x \vee y]$. Hence, $z$ is the difference "$x-y$". By $c')$ we have $y \in I$ or $z \in I$. If $y \in I$, we are done. If $z \in I$, then $z \vee (x \wedge y) \in I$ because $I$ is an ideal. But 
$$z \vee (x \wedge y)= (z \vee x) \wedge (z \vee y)=x \wedge (x \vee y)=x$$ 
We used $z \vee x=x$. That follows from $z= z \wedge (x \vee y)=(z \wedge x ) \vee (z \wedge y)= (z \wedge x ) \vee 0= z \wedge x$, which means $z \leq x$.
