Consider
.. X = {1, 2, 3, 4}, x = 4
.. Y = {2, 3, 4, 5, 6}, y = 5, w = 3
The elements in X or Y (X ∪ Y) are {1, 2, 3, 4, 5, 6}, n = 6.
.. n = 6 = 4 + 5 - 3
Note that if we just add x and y, we count the common elements twice. In order to just count the common elements once, we need to subtract that count from the total of x and y.
selection B is appropriate.
Answer:
The value of the proposition is FALSE
Step-by-step explanation:
~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]
Let's start with the smallest part: ~X. The symbol ~ is negation when X is true with the negation is false and vice-versa. In this case, ~X is true (T)
~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ T) v (B ⊃ X)]
Now the parts inside parenthesis: (A ⊃ Y),(X ⊃ B),(A ≡ T) and (B ⊃ X). The symbol ⊃ is the conditional and A ⊃ Y is false when Y is false and A is true, in any other case is true. The symbol ≡ is the biconditional and A ≡ Y is true when both A and Y are true or when both are false.
(A ⊃ Y) is False (F)
(X ⊃ B) is True (T)
(A ≡ T) is True (T)
(B ⊃ X) is False (F)
~[(F) v ~(T)] ⋅ [~(T) v (F)]
The two negations inside the brackets must be taken into account:
~[(F) v F] ⋅ [F v (F)]
The symbol left inside the brackets v is the disjunction, and A v Y is false only with both are false. F v (F) is False.
~[F] ⋅ [F]
Again considerating the negation:
T⋅ [F]
Finally, the symbol ⋅ is the conjunction, and A v Y is true only with both are true.
T⋅ [F] is False.
Answer:
13 sweets
Step-by-step explanation:
Fiona : Angad
2 : 1
26 : X
26/2 = X/1
X = 13