To show that ( A \oplus B = (A \cup B) - (A \cap B) ), we need to prove two inclusions.
For the first inclusion, let ( x \in A \oplus B ). This means that ( x ) is in exactly one of ( A ) or ( B ), but not both. Therefore, ( x ) is in ( A ) or ( B ), but not in their intersection. Hence, ( x \in (A \cup B) - (A \cap B) ).
For the second inclusion, let ( x \in (A \cup B) - (A \cap B) ). This means that ( x ) is in either ( A ) or ( B ), but not in their intersection. Thus, ( x ) is in exactly one of ( A ) or ( B ), leading to ( x \in A \oplus B ).
Therefore, we have shown that ( A \oplus B = (A \cup B) - (A \cap B) ).
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
all major intersection have various other signs offering information to road users
The set A union B can be decomposed into three disjoint sub sets A\ (A int B), B\(A int B), and (A int B). So in this case (A union B) and (A int B) are fixed but "moving" elelments from A\ (A int B) into B\(A int B) will not affect (A union B) and (A int B). You should be able to fill in the details now.
a XOR bis equivalent to: (a AND NOT b) OR (b AND NOT a)
a XOR b is a^b in C language
Draw your Venn Diagram as three overlapping circles. Each circle is a set. The union of the sets is what's contained within all 3 circles, making sure not to count the overlapping portion twice. An easier problem is when you have 2 sets, lets say A and B. In a Venn Diagram that looks like 2 overlapping circles. A union B = A + B - (A intersect B) A intersect B is the region that both circles have in common. You subtract that because it has already been included when you added circle A, so you don't want to add that Again with circle B, thus you subtract after adding B. With three sets, A, B, C A union B union C = A + B - (A intersect B) + C - (A intersect C) - (B intersect C) + (A intersect B intersect C) You have to add the middle region (A intersect B intersect C) back because when you subtract A intersect C and B intersect C you are actually subtracting the very middle region Twice, and that's not accurate. This would be easier to explain if we could actually draw circles.
A = A xor B B = A xor B A = A xor B in C... A^=B; B^=A; A^=B;
a := a XOR b b := a XOR b a := a XOR b it works, but never use it in real programs do you know why its not used in real program??
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
Non-using of temporal variable. It is possible, but has no use in real programming. Eg. a := a XOR b b := a XOR b a := a XOR b I warn you again: this is just a curiosity, not something you really use.
In some cases, A union B is convex, but in general this may not be true. Consider two sets A, B (subsets of Rn) such that A intersect B is the null set. Now choose a point x in A, and y in B. If a set is to be convex, then all points on the line tx + (1-t)y (0
For 2-input EX-OR gate, if one input is A, the other input is B, and the output is Y. Then the Boolean expression for EX-OR (XOR) function (gate) is Y=A⊕B The output Y is true if either input A or if input B is true, but not both.Y= ( (A and NOT B) or (NOT A and B) ) ;
void swap (int *a, int *b) { *a ^= *b; *b ^= *a; *a ^= *b; return; }
Let x be in A intersect B. Then x is in A and x is in B. Then x is in A.
all major intersection have various other signs offering information to road users
// Note: ^ is the XOR operator a = a ^ b b = b ^ a a = a ^ b