Take a two-stage game with complete information and simultaneous actions in each state:
(1) Player 1 and 2 simultaneously choose action $a_1\in A_1$ and $a_2\in A_2$ respectively.
(2) Player 1 and 2 observe the outcome of the 1st stage $(a_1, a_2)$, then simultaneously choose action $a_3\in A_3$ and $a_4\in A_4$ respectively.
Payoffs are $u_i(a_1, a_2, a_3, a_4)$ for $i = 1,2$.
Suppose that I do the following (typically called backward induction):
(A) find the functions $a^*_3(a_1,a_2)$ and $a^*_4(a_1,a_2)$ such that
$$ \begin{cases} a_3^*(a_1, a_2)\in argmax_{a_3(\cdot)}u_1(a_1, a_2, a_3(a_1, a_2), a_4(a_1, a_2))\\ a_4^*(a_1, a_2)\in argmax_{a_4(\cdot)}u_2(a_1, a_2, a_3(a_1, a_2), a_4(a_1, a_2))\\ \end{cases} $$
(B) find $a_1^*, a_2^*$ such that $$ \begin{cases} a_1^*\in argmax_{a_1}u_1(a_1, a_2, a^*_3(a_1, a_2), a^*_4(a_1, a_2))\\ a_2^*\in argmax_{a_2}u_2(a_1, a_2, a^*_3(a_1, a_2), a^*_4(a_1, a_2))\\ \end{cases} $$
Which equilibrium notion I'm I applying?
Basically, in (A) I find the pure strategy Nash Equilibrium of stage 2, for each possible action played in stage 1; in (B) I find the pure strategy Nash Equilibrium of stage 1, given the solution from (A).
How is $\{a_1^*, a_2^*, a^*_3(a_1, a_2), a^*_4(a_1, a_2)\}$ called?
