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Given State Table
q | x=0 x=1 | z
--------------------------------
A | C D | 1
B | C D | 0
C | B D | 1
D | C A | 1
Step 1 produces five SP Partitions
P1 = (AB)(C)(D)
P2 = (ABC)(D)
P3 = (AD)(B)(C)
P4 = (A)(BC)(D)
P5 = (ABD)(C)
Step 2 requires three sums
P1 + P3 = (ABD)(C)--> P5
P1 + P4 = (ABC)(D)--> P2
P3 + P4 = (AD)(BC)--> P6
only one new partition is found by step 2.
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