If a ∧ b is a contradiction, P(a | e) + P(b | e) = P(a ∨ b | e)是什么意思
时间: 2024-03-18 11:45:51 浏览: 17
这个问题涉及到概率和逻辑的知识。
首先,"a ∧ b is a contradiction"表示a和b是互相矛盾的,即它们不能同时成立。
接着,"P(a | e)"表示在已知条件e下,a成立的概率;"P(b | e)"表示在已知条件e下,b成立的概率;"P(a ∨ b | e)"表示在已知条件e下,a或b成立的概率。
那么,根据概率的加法规则,对于任意两个事件X和Y,P(X ∨ Y) = P(X) + P(Y) - P(X ∧ Y)。因为a和b是互相矛盾的,所以P(a ∧ b) = 0。因此,P(a ∨ b) = P(a) + P(b)。
将这个公式代入原式中,得到 P(a | e) + P(b | e) = P(a) + P(b)。
这个公式的意义是,在已知条件e下,a和b的概率之和等于a或b的概率。这个公式也可以被解释为,如果a和b是互相矛盾的,那么在已知条件e下,a和b不能同时成立,因此它们的概率之和等于a或b成立的概率。
相关问题
We show that every atom a that is added to C by the algorithm is a logical consequence of KB ∪ Percepts. Suppose there is an atom h ∈ C that is not a logical consequence. Let h be the first ever such atom to be added in C. There must be some clause in KB ∪ Percepts, in the form h ← a1 ∧ · · · ∧ am such that a1, . . . , am are all in C. By assumption, KB ∪ Percepts |= ai for all 1 ≤ i ≤ m. Then it must be that KB ∪ Percepts |= h. Contradiction.是什么意思
这段文字描述了一个关于知识库(KB)和感知(Percepts)的算法,在该算法中,每个添加到集合C中的原子a都是KB和Percepts的逻辑结果。如果有一个原子h在C中,但不是它们的逻辑结果,那么假设h是第一个这样的原子被添加到C中,那么一定存在一个KB和Percepts中的子句,形式为h ← a1 ∧ · · · ∧ am,其中a1,...,am都在C中。由于假设KB和Percepts对于1 ≤ i ≤ m中的所有ai都成立,因此KB和Percepts也应该对h成立。但是这与假设相矛盾,即h不是KB和Percepts的逻辑结果。因此,原命题的结论是每个添加到C中的原子a都是KB和Percepts的逻辑结果。
Show that the columns of X are linearly independent if the constraint α1 = 0 is imposed.
To show that the columns of X are linearly independent if the constraint α1 = 0 is imposed, we need to show that the only solution to the equation Xα = 0 (where α is a vector of coefficients) is the trivial solution α = 0.
Let X be an n x p matrix with columns x1, x2, ..., xp. If we impose the constraint α1 = 0, then the equation Xα = 0 becomes:
0*x1 + α2*x2 + α3*x3 + ... + αp*xp = 0
This can be rewritten as:
α2*x2 + α3*x3 + ... + αp*xp = 0
Assume that the columns of X are linearly dependent, so there exist coefficients α2, α3, ..., αp, not all zero, such that α2*x2 + α3*x3 + ... + αp*xp = 0. Without loss of generality, we can assume that α2 ≠ 0.
Then we can solve for x2 in terms of x3, ..., xp as:
x2 = (-α3/α2)*x3 - ... - (αp/α2)*xp
Substituting this expression into the equation α2*x2 + α3*x3 + ... + αp*xp = 0, we get:
α2*(-α3/α2)*x3 - α2*(α4/α2)*x4 - ... - α2*(αp/α2)*xp + α3*x3 + ... + αp*xp = 0
Simplifying this expression, we get:
-α3*x3 - α4*x4 - ... - αp*xp + α3*x3 + ... + αp*xp = 0
This is a contradiction, since we assumed that not all of the coefficients α2, α3, ..., αp are zero, but we ended up with the equation 0 = 0. Therefore, our assumption that the columns of X are linearly dependent must be false, and we can conclude that the columns of X are linearly independent if the constraint α1 = 0 is imposed.
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