For j =1 to 6
zj = xj ^ yj
Split equations. For {zj} we get well known system
Solutions standard bitmask table for z6,z5,z4,z3,z2,z1.
z2 -> z1 =1
z3 -> z2 =1
z4 -> z3 =1
z5 -> z4 =1
z6 -> z5 =1
z6 z5 z4 z3 z2 z1
----------------------------------
1 1 1 1 1 1
0 1 1 1 1 1
0 0 1 1 1 1
0 0 0 1 1 1
0 0 0 0 1 1
0 0 0 0 0 1
0 0 0 0 0 0
1. Consider zj = 0 then xj ^ yj = 0
So ( xj=1, yj=0); (xj=0, yj=1) ; (xj=0, yj=0)
Look back into original system. For each j we should have
(xj v yj) = 1 hence decline third pair.
Total 2 pairs for zj = 0
2. Consider zj = 1 then xj ^ yj =1
So xj=1 and yj=1
Just one solution.
Thus we get :-
z6 z5 z4 z3 z2 z1
----------------------------------
1 1 1 1 1 1 1
0 1 1 1 1 1 2^1
0 0 1 1 1 1 2^2
0 0 0 1 1 1 2^3
0 0 0 0 1 1 2^4
0 0 0 0 0 1 2^5
0 0 0 0 0 0 2^6
Total (2^7 -1)/(2-1) = 127
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Mapping method
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(x1 v y1)^((x2^y2) => (x1^y1)) = 1
(x2 v y2)^((x3^y3) => (x2^y2)) = 1
. . . . .
(x5 v y5)^((x6^y6) => (x5^y5)) = 1
(x6 v y6) = 1
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