In the next step, backward implication of the gate at J-frontier (gate-3) implies net j =1 and primary input d =1. This is illustrated in Figure 7(g).

 Figure 7(g): Backward implication

Assignment of net j =1 is inconsistency because it implies output of gate-1 to be 1, whereas it needs to be D (for fault sensitization). Inconsistency is shown in Figure 7(h)
 

Figure 7(h): Inconsistency in net j

After this inconsistency we need to backtrack. To backtrack we need to find points were there were choices and we took one of them. In this example, there was a choice of selecting one gate (for propagating D) among two gates in the D-frontier; see Figure 7(d), where gate-2 was selected. Now let us backtrack by taking the other choice, i.e., consider gate-3 for D propagation; this is shown in Figure 7(i). It may be noted that for backtracking we have to undo all steps starting from where we got inconsistency to the point where another choice is to be tried. In Figure 7(i) all steps are undone (by making net values to X).

 Figure 7(i): Backtrack by considering gate-3 for D propagation

Similar to the step in Figure 7(e), we propagate D though gate-3 (the other choice). This makes net k =  by forward implication and d =1 by backward implication. The step is illustrated in Figure 7(j). After the propagation of   though gate-2, the new D-frontier comprises gate-4.
 Figure 7(j): Propagation of D and implications