To find all subgraphs of $G$ of order 4 and size 4, we need to look for subgraphs that have exactly 4 vertices and 4 edges.

There are ${6 \choose 4} = 15$ ways to choose 4 vertices from $V$. We will consider each of these sets of 4 vertices and check if they form a subgraph of order 4 and size 4.

The possible sets of vertices are:

• ${a,b,c,d}$
• ${a,b,c,e}$
• ${a,b,c,f}$
• ${a,b,d,e}$
• ${a,b,d,f}$
• ${a,b,e,f}$
• ${a,c,d,e}$
• ${a,c,d,f}$
• ${a,c,e,f}$
• ${a,d,e,f}$
• ${b,c,d,e}$
• ${b,c,d,f}$
• ${b,c,e,f}$
• ${b,d,e,f}$
• ${c,d,e,f}$

Now, for each set of 4 vertices, we will check if there are 4 edges connecting them. If so, then this set of vertices forms a subgraph of order 4 and size 4.

• ${a,b,c,d}$: This subgraph has edges $ab$, $ad$, $bd$, and $cd$. So it is a subgraph of order 4 and size 4.
• ${a,b,c,e}$: This subgraph has edges $ab$, $be$, $ac$, and $bc$. So it is not a subgraph of order 4 and size 4.
• ${a,b,c,f}$: This subgraph has edges $ab$, $af$, $ac$, and $bc$. So it is not a subgraph of order 4 and size 4.
• ${a,b,d,e}$: This subgraph has edges $ab$, $ae$, $bd$, and $de$. So it is a subgraph of order 4 and size 4.
• ${a,b,d,f}$: This subgraph has edges $ab$, $af$, $bd$, and $df$. So it is not a subgraph of order 4 and size 4.
• ${a,b,e,f}$: This subgraph has edges $ab$, $af$, $be$, and $ef$. So it is not a subgraph of order 4 and size 4.
• ${a,c,d,e}$: This subgraph has edges $ad$, $ae$, $cd$, and $ce$. So it is a subgraph of order 4 and size 4.
• ${a,c,d,f}$: This subgraph has edges $ad$, $af$, $cd$, and $cf$. So it is not a subgraph of order 4 and size 4.
• ${a,c,e,f}$: This subgraph has edges $ac$, $af$, $ce$, and $cf$. So it is not a subgraph of order 4 and size 4.
• ${a,d,e,f}$: This subgraph has edges $ad$, $ae$, $df$, and $ef$. So it is a subgraph of order 4 and size 4.
• ${b,c,d,e}$: This subgraph has edges $bd$, $be$, $cd$, and $ce$. So it is a subgraph of order 4 and size 4

The complement of $G$, denoted $G^c$, is a graph with the same vertex set $V$ as $G$, but where two vertices are adjacent in $G^c$ if and only if they are not adjacent in $G$. Therefore, the number of edges in $G^c$ is given by:

$$|E(G^c)|=\binom{n}{2}-|E(G)|$$

where $\binom{n}{2}$ is the total number of possible edges in a graph of order $n$. So the order of $G^c$ is $n$ and its size is $\binom{n}{2}-m$.

The graph obtained from $G$ by deleting vertex $v$, denoted $G-v$, has the same vertex set as $G$ but with $v$ removed. Therefore, the order of $G-v$ is $n-1$. To find the size of $G-v$, we need to count the number of edges in $G$ that do not contain $v$. Let $m_v$ denote the number of edges incident to $v$. Then, the size of $G-v$ is given by:

$$|E(G-v)|=m-m_v$$

The graph obtained from $G$ by deleting edge $e$, denoted $G-e$, has the same vertex set as $G$ but with $e$ removed. Therefore, the order of $G-e$ is $n$. To find the size of $G-e$, we simply subtract 1 from the size of $G$:

$$|E(G-e)|=m-1$$

since we are removing a single edge.

Let $G=(X,Y;E)$ be a bipartite graph of order $n$, with $|X|=a$ and $|Y|=b$. Then the size of the graph, denoted $|E|$, is given by:

$$|E|=\sum_{x\in X}d(x) = \sum_{y\in Y}d(y)$$

where $d(x)$ denotes the degree of vertex $x$ in $G$. Since $G$ is bipartite, we have $d(x)=|N(x)|$, where $N(x)$ denotes the set of neighbors of $x$ in $G$. Furthermore, we have $N(x)\subseteq Y$ for all $x\in X$, and $N(y)\subseteq X$ for all $y\in Y$.

Now, we use the inequality $|N(x)|\leq b$ for all $x\in X$ (since $N(x)\subseteq Y$), and similarly $|N(y)|\leq a$ for all $y\in Y$ (since $N(y)\subseteq X$). Therefore:

$$|E|=\sum_{x\in X}d(x) = \sum_{x\in X}|N(x)|\leq \sum_{x\in X} b = ab = \frac{n^2}{4}$$

where the last equality follows from $a+b=n$ and the fact that the maximum value of $ab$ is attained when $a=b=n/2$. Similarly, we have:

$$|E|=\sum_{y\in Y}d(y) = \sum_{y\in Y}|N(y)|\leq \sum_{y\in Y} a = ab = \frac{n^2}{4}$$

Therefore, we have shown that $|E|\leq n^2/4$, as required.