A. For all integers a and b let a+b is odd. If a and b are both even than ∃p,l∈Za=2p,b=2l⟹a+b=2(p+l) is even too. If a and b are both odd than ∃p,l∈Za=2p+1,b=2l+1⟹a+b=2(p+l+1) is even too. So we have only one of a and b is odd and ∃p,l∈Za=2p,b=2l+1⟹a+b=2(p+l)+1 .
B. For any integer n the number (n3−n) equals to n(n+1)(n−1) that is the product of three three consecutive natural numbers, one of which is even. In case (n−1)=0 we have (n3−n)=0 and it is even too.
C. Let's prove that for alll sets A and B the equality ¬(A∧B)=¬A∨¬B .
¬(A∧B)=(¬A∧B)∨(A∧¬B)∨(¬A∧¬B)=
((¬A∧B)∨(¬A∧¬B))∨((A∧¬B)∨(¬A∧¬B))=
(¬A∧(B∨¬B))∨(¬B∧(B∨¬B))=¬A∨¬B
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