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Given State Table
q | x=0 x=1 | z
--------------------------------
A | C D | 0
B | D A | 0
C | E D | 0
D | B A | 1
E | C D | 1
Step 1 produces five SP Partitions
P1 = (ACE)(BD)
P2 = (ADE)(BC)
P3 = (AE)(B)(C)(D)
P4 = (A)(BD)(C)(E)
P5 = (A)(B)(CE)(D)
Step 2 requires three sums
P1 + P4 = (ACE)(BD)--> P6
P3 + P4 = (AE)(BD)(C)--> P7
P4 + P5 = (A)(CE)(BD)--> P8
P7 + P8 = (ACE)(BD)--> P6
There are eight non trivial SP partition, of which two are two block and none are output consistent
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