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+//Example 9.12: Reduction of state table
+clc // Clears the console
+disp("Given State Table")
+disp("q | x=0 x=1 | z1 z2 z3 z4 z5")
+disp('--------------------------------')
+disp("A | D B | 0 0 0 1 1")
+disp("B | E C | 0 0 1 0 1")
+disp("C | A B | 1 1 0 0 1")
+disp("D | E C | 1 1 1 1 1")
+disp("E | D B | 1 0 0 1 1")
+disp('Step 1 produces given SP Partitions')
+disp('P1 = (ADE)(BC)')
+disp('P2 = (AE)(B)(C)(D)')
+disp('P3 = (AE)(BC)(D)')
+disp('P4 = (A)(BD)(C)(E)')
+disp('P5 = (AE)(BCD)')
+disp('Step 2 requires three sums')
+disp('P2 + P4 = (AE)(BD)(C)--> P6')
+disp('There are six non trivial SP partitions.')
+disp('For the first output column, None of the SP partitions are output consistent')
+disp('for the second output column only P2 is output consistent')
+disp("q | x=0 x=1 | z2")
+disp('--------------------------------')
+disp("A | D B | 0")
+disp("B | A C | 0")
+disp("C | A B | 1")
+disp("D | A C | 1")
+disp('for the third output column only P2, P4 and P6 all are output consistent')
+disp("q | x=0 x=1 | z3")
+disp('--------------------------------')
+disp("A | B B | 0")
+disp("B | A C | 0")
+disp("C | A B | 1")
+disp('for the fourth output column only P1 is output consistent')
+disp("q | x=0 x=1 | z3")
+disp('--------------------------------')
+disp("A | A B | 0")
+disp("B | A B | 1")
+disp('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')