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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
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