The $E$ theorem reaches an induced $P_5$ witness

The identity map on $\{0,1,2,3,4\}$ preserves adjacency and nonadjacency, so it is an induced embedding of $P_5$ in itself by [L3]; hence $P_5$ is not $P_5$-free.

The six-element set $\{p_1,\dots ,p_5,q\}$ has no injection into the five-element set $\{0,1,2,3,4\}$; hence no induced embedding of $E$ in $P_5$ exists, and $P_5$ is $E$-free.

Suppose the finite graph $G$ is not $E$-free. By [L3] there is an induced embedding $\phi$ of $E$ in $G$. Its restriction ${\phi }^{\mathrm{\prime }}$ to $\{p_1,\dots ,p_5\}$ is again an injection preserving adjacency and nonadjacency, and by [L1] the induced subgraph of $E$ on those five vertices has exactly the edges $p_1{p}_2,p_2{p}_3,p_3{p}_4,p_4{p}_5$, which is a $P_5$ under [L2]. So ${\phi }^{\mathrm{\prime }}$ is an induced embedding of $P_5$ in $G$, and $G$ is not $P_5$-free. Contrapositively, every $P_5$-free graph is $E$-free.

By step 1.3 the class of $P_5$-free graphs is contained in the class of $E$-free graphs, and the containment is strict because the graph $P_5$ itself is $E$-free by step 1.2 yet is not $P_5$-free by step 1.1.

The $E$ theorem [L4] bounds every nonempty graph in the larger $E$-free class, hence in particular every nonempty graph in the $P_5$-free class, whereas the earlier $P_5$ theorem [L5] concerns only that smaller class. Since both theorems merely assert the existence of positive exponents, step 2.1 is a strict inclusion of hypothesis classes and implies no comparison between the two exponents.

The witness $P_5$, the general inclusion of step 1.3 and the strictness of step 2.1 verify all assertions of the example.