Solution to A- For all integers a and b, if a + b is odd, then a … - Sikademy
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Archangel Macsika

A- For all integers a and b, if a + b is odd, then a is odd or b is odd. B- For any integer n the number (n3 - n) is even. C- Proof of De-Morgan’s Law

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A. For all integers a  and b let a+b is odd. If a and b are both even than \exist p, l \in Z a=2p, b=2l\implies a+b=2(p+l) is even too. If a and b are both odd than \exist p, l \in Z a=2p+1, b=2l+1 \implies a+b=2(p+l+1) is even too. So we have only one of  a  and b is odd and \exist p, l \in Z a=2p, b=2l+1 \implies a+b=2(p+l)+1 .

B. For any integer n  the number (n^3-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 (n^3-n)=0 and it is even too.

C. Let's prove that for alll sets A and B the equality \lnot(A\land B)=\lnot A\lor \lnot B .

\lnot (A\land B)=(\lnot A\land B)\lor ( A\land \lnot B) \lor ( \lnot A\land \lnot B) =

((\lnot A\land B) \lor ( \lnot A\land \lnot B))\lor (( A\land \lnot B) \lor ( \lnot A\land \lnot B)) =

( \lnot A \land ( B\lor \lnot B))\lor (\lnot B \land ( B\lor \lnot B)) =\lnot A\lor \lnot B

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