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From paper 2019.02.13 Problem 2(a)

How did we go from (!b & c ^c ) to (b&c)?Since and binds stronger than XOR.

I solved it as (!b & c ^c ) (distribution) -> (!b & c) ^ (!b&c) -> 0. T^T or F^F = 0

!b&c xor c = (c ? !b&1 xor 1 : !b&0 xor 0) = (c ? !b xor 1 : 0 xor 0) = (c ? b : 0) = b&c