Here, we have,
$G_1 = (X_{1a} \oplus X_{2a}) \vee (X_{1b} \oplus X_{2b}) $
$G_2 = (X_{2a} \oplus X_{3a}) \vee (X_{2b} \oplus X_{3b}) $
$G_3 = (X_{3a} \oplus X_{4a}) \vee (X_{3b} \oplus X_{4b}) $
$G_4 = (X_{4a} \oplus X_{5a}) \vee (X_{4b} \oplus X_{5b}) $
$G_5 = (X_{5a} \oplus X_{1a}) \vee (X_{5b} \oplus X_{1b}) $
$F=(X_{1a} \vee X_{1b} ) \wedge (X_{2a} \vee X_{2b} ) \wedge (X_{3a} \vee X_{3b} ) \wedge (X_{4a} \vee X_{4b} ) \wedge (X_{5a} \vee X_{5b} )$
$H= F \wedge G_1 \wedge G_2 \wedge G_3 \wedge G_4 \wedge G_5 $
Now, we have to assign boolean variables $X_{1a}$ to $X_{5a}$ and $X_{1b}$ to $X_{5b}$ as $1$ or $0$ in such a way that $H$ becomes $1$
To make $H$ as $1$, $F$ and all $G_1$ to $G_5$ should be $1.$ Now, either we start with $G_1$ or $G_2 $ or any $G_i, 1\leq i \leq 5 $, answer will remain same.
So, first to make both $G_1$ and $F$ as $1$ , possibilities are :
$(i)$ $X_{1a} = 1$ and $X_{1b} = 1$
$(ii)$ $X_{1a} = 1$ and $X_{1b} = 0$
$(iii)$ $X_{1a} = 0$ and $X_{1b} = 1$
So, case $(i)$ $X_{1a} = 1$ and $X_{1b} = 1$
Now, to make $F=1$ and $G_1 = 1$ $\Rightarrow$ $(1 \oplus X_{2a}) \vee (1 \oplus X_{2b}) =1 $ means both $(1 \oplus X_{2a})$ and $(1 \oplus X_{2b}) $ should not be $0$. As we know, that $x \oplus x = 0.$. So, $X_{2a} \neq 1$ and $X_{2b} \neq 1$.
So, when $X_{1a} = 1$ and $X_{1b} = 1$,
$X_{2a}=1, X_{2b}=0$
$X_{2a}=0, X_{2b}=1$.
So, $(1,1)$ maps to either $(1,0)$ or $(0,1)$ i.e.
$(1,1) \mapsto (1,0)$ (or) $(1,1) \mapsto (0,1)$
Similarly,
To make $F=1$ and $G_2 = 1$ $\Rightarrow$ $(1 \oplus X_{3a}) \vee (0 \oplus X_{3b}) =1 $ (or) $(0 \oplus X_{3a}) \vee (1 \oplus X_{3b}) =1 $ means both $(1 \oplus X_{3a})$ and $(0 \oplus X_{3b}) $ should not be $0$ and . As we know, that $x \oplus x = 0.$. So, $X_{3a} \neq 1$ and $X_{3b} \neq 0$ (or) $X_{3a} \neq 0$ and $X_{3b} \neq 1$.
So, when $X_{2a}=1, X_{2b}=0$ then $X_{3a}=0,$ $X_{3b}=1$ (or) $X_{3a}=1,$ $X_{3b}=1$ and
when $X_{2a}=0, X_{2b}=1$ then $X_{3a}=1,$ $X_{3b}=0$ (or) $X_{3a}=1,$ $X_{3b}=1$
So, $(1,0) \mapsto (0,1)$ (or) $(1,0) \mapsto (1,1)$ and
$(0,1) \mapsto (1,0)$ (or) $(0,1) \mapsto (1,1).$
Like this we can find the values of other boolean variables very easily if we know the following mappings :
Using these mappings, we can get the possible values of other boolean variables.
So, here, $(X_{ia},Y_{ib})$ represents the value of boolean variables $X_{ia}$ and $X_{ib}$ at level $i$ where $1 \leq i \leq 5$
Now, $G_5$ will be $0$ when $(X_{1a},X_{1b}) \oplus (X_{5a},X_{5b}) = 0$ which says those leaf nodes which have value same as $(1,1)$, $G_5$ will be zero. So, we have to exclude $(1,1)$ from leaf nodes. Out of $16$ leaf nodes, $6$ has values as $(1,1)$. So, remaining values will be $16-6 = 10.$
So, when $X_{1a} = 1$ and $X_{1b} = 1$, $H=1$ for $10$ assignments.
Now for case $(ii) X_{1a} = 1$ and $X_{1b} = 0$
Tree structure will be like :
Here, total $6$ $(1,0)$(starting node) in leaf nodes. So, total $16-6 =10$ assignments for which $H=1$
Now, for case $(iii) X_{1a} = 0$ and $X_{1b} = 1$
Tree structure will be like :
Here, also $6$ $(0,1)$(starting node) in leaf nodes. So, total $16-6 =10$ assignments for which $H=1$
So, Total assignments in all 3 cases = $10+10+10 = 30.$
Now, since, structure is symmetric, so, you can start with either $(X_{2a},X_{2b})$ or $(X_{3a},X_{3b})$ or take any case, answer will remain the same.