Given State Table

 q |    x=0     x=1   |    z1   z2   z3   z4   z5

 --------------------------------

 A |     D       B    |    0    0    0    1    1

 B |     E       C    |    0    0    1    0    1

 C |     A       B    |    1    1    0    0    1

 D |     E       C    |    1    1    1    1    1

 E |     D       B    |    1    0    0    1    1

 Step 1 produces given SP Partitions

 P1 = (ADE)(BC)

 P2 = (AE)(B)(C)(D)

 P3 = (AE)(BC)(D)

 P4 = (A)(BD)(C)(E)

 P5 = (AE)(BCD)

 Step 2 requires three sums

 P2 + P4 = (AE)(BD)(C)--> P6

 There are six non trivial SP partitions.

 For the first output column, None of the SP partitions are output consistent

 for the second output column only P2 is output consistent

 q |    x=0         x=1   |    z2

 --------------------------------

 A |     D           B    |    0

 B |     A           C    |    0

 C |     A           B    |    1

 D |     A           C    |    1

 for the third output column only P2, P4 and P6 all are output consistent

 q |    x=0         x=1   |    z3

 --------------------------------

 A |     B           B    |    0

 B |     A           C    |    0

 C |     A           B    |    1

 for the fourth output column only P1 is output consistent

 q |    x=0         x=1   |    z3

 --------------------------------

 A |     A           B    |    0

 B |     A           B    |    1

 for the last output column there is no need to find the SP partitions, The system is combinational. It does not depend on the state z=1