 The graph is the same as that shown in  Figure 4  above. Again, we want to show that \(S = 1\) is an actual cause of \(W = 1\). Conditions AC(1) and AC(2) are obviously satisfied. For AC(3), we choose \(\bX = \{B, S\}\) and \(\bZ = \varnothing\). For condition AC(3a), we choose for our alternate setting \(\bX = \bx'\) \(B = 0\) and \(S = 0\). Once again, the counterfactual \({S = 0} \amp {B = 0} \boxright {W = 0}\) is true. Now, for AC(3b) we must show that \(\bX = \{B, S\}\) is minimal. It is easy to check that \(\{B\}\) alone won’t satisfy AC(3a). Whether we take \(\bZ = \varnothing\) or \(\bZ = \{S\}\), changing B to 0 (perhaps while also setting S to 1) will not change the value of W. A parallel argument shows that \(\{S\}\) alone won’t satisfy AC(3a) either. The key idea here is that S is a member of a minimal set of variables that need to be changed in order to change the value of W.