(∀x)((∃y)(A(x, y)∧B(y))→(∃y)(C(y)∧D(x, y)))子集化简

这个式子可以进行子集化简，具体步骤如下：

1. 将条件语句符号化，即将箭头变成逻辑或符号和逻辑非符号：

(∀x)((∃y)(A(x, y)∧B(y))∧¬(∃y)(C(y)∧D(x, y)))∨(∃y)(C(y)∧D(x, y))

2. 对于第一个子句使用德摩根定律将否定符号移入括号内并对括号内的量词进行调换：

(∀x)(∀y)(A(x, y)∧B(y)→(C(y)∧¬D(x, y)))∨(∃y)(C(y)∧D(x, y))

3. 将第一个子句中的条件语句符号化，即将箭头变成逻辑或符号和逻辑非符号：

(∀x)(∀y)(¬A(x, y)∨¬B(y)∨C(y)∧¬D(x, y)))∨(∃y)(C(y)∧D(x, y))

4. 对于第一个子句使用德摩根定律将否定符号移入括号内并对括号内的量词进行调换：

(∀x)(∃y)(A(x, y)∧B(y)∧(¬C(y)∨D(x, y)))∨(∃y)(C(y)∧D(x, y))

这样就完成了子集化简。

相关问题

Let x, y ∈ R, show that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0)

To prove that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0), we need to prove both directions of the equivalence:

First, let's prove that if ∀x∃y(xy = 0), then ∃y∀x(xy = 0):

Assume ∀x∃y(xy = 0). We want to show that there exists a y such that for all x, xy = 0.

Let's choose an arbitrary value of y such that for any x, either xy = 0 (if x ≠ 0) or y = 0 (if x = 0). Such a value of y exists since for any nonzero x, there exists a y = 0 such that xy = 0, and for x = 0, any y would satisfy xy = 0.

Now, we need to show that this y satisfies ∀x(xy = 0):

Take an arbitrary x. If x = 0, then xy = 0 (since y can be any value). If x ≠ 0, then y was chosen so that xy = 0. Therefore, in either case, xy = 0, and we have shown that ∃y∀x(xy = 0).

Now let's prove the other direction, that if ∃y∀x(xy = 0), then ∀x∃y(xy = 0):

Assume ∃y∀x(xy = 0). We want to show that for any x, there exists a y such that xy = 0.

Take an arbitrary x. Since ∃y∀x(xy = 0), there exists a y such that for all x, xy = 0. In particular, this y satisfies xy = 0 for our chosen value of x. Therefore, we have shown that for any x, there exists a y such that xy = 0, and thus we have proved ∀x∃y(xy = 0).

Since we have proved both directions of the equivalence, we can conclude that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0).

∀xF(x)→∃yG(y) 与 ∀x∃y(F(x)→G(y))是否等值

这两个命题不等值。

首先，对于前者，对于任何一个x使得F(x)成立，那么根据蕴含的定义，F(x)→∃yG(y)也必定成立。因此，我们可以得出结论：如果存在一个x使得F(x)成立，那么∃yG(y)也必定成立。也就是说，前者的真值只与是否存在一个满足F(x)的x有关。

而对于后者，我们可以发现，对于任何一个x，如果F(x)成立，那么根据蕴含的定义，F(x)→G(y)也必定成立，其中y可以是任何一个使得G(y)成立的值。因此，我们可以得出结论：如果存在一个x使得F(x)成立，那么∃y(F(x)→G(y))也必定成立。也就是说，后者的真值不仅与是否存在一个满足F(x)的x有关，还与是否存在一个y使得G(y)成立有关。

因此，这两个命题不等值。