(q∧r∧p)∨(q∧r∧¬p)化简成qrp形式

(q∧r∧p)∨(q∧r∧¬p) 可以使用分配律和合取范式进行化简：

(q∧r∧p)∨(q∧r∧¬p)

= (q∧r∧(p∨¬p)) ∨ (q∧r∧¬(p∧¬p)) （分配律）

= (q∧r∧1) ∨ (q∧r∧0) （p∨¬p = 1，p∧¬p = 0）

= qrp （合取范式）

因此，原命题可以化简为 qrp 的形式。

相关问题

(p→((¬r∧s)∨(r∧¬s)))∧(¬r∧q)∨(r∧¬q)∧(r→¬s)求主析取范式的详细过程

首先，我们需要将条件式转化为合取范式，即将它们转换为逻辑与的形式。具体步骤如下：

1. 将条件式中的蕴含符号转换为逻辑或和非运算的组合形式，即 p→q 转换为 ¬p∨q。

(p→((¬r∧s)∨(r∧¬s)))∧(¬r∧q)∨(r∧¬q)∧(r→¬s)
= (¬p∨((¬r∧s)∨(r∧¬s)))∧(¬r∧q)∨(r∧¬q)∧(¬r∨¬s)

2. 将条件式中的合取符号转换为逻辑或和非运算的组合形式，即 p∧q 转换为 ¬(¬p∨¬q)。

= ((¬¬p∧¬(¬r∧s)∧¬(r∧¬s))∨(¬¬p∧(¬r∧s)∨¬¬p∧(r∧¬s)))∧(¬¬r∨¬q)∨(r∧¬q)∧(¬¬r∨¬¬s)

3. 对条件式中的每个逻辑非运算符号应用德·摩根定律，并将条件式中的双重否定符号去除，即 ¬¬p 转换为 p。

= ((p∨(r∨¬s))∧(p∨(¬r∨s)))∧(r∨q)∧(¬r∨¬q)∧(r∨¬q)∧(¬r∨s)

4. 对条件式中的每个逻辑与运算符号应用分配律。

= ((p∧p)∨(p∧¬r)∨(p∧s)∨(r∧p)∨(r∧¬r)∨(r∧s))∧(r∨q)∧(¬r∨¬q)∧(r∨¬q)∧(¬r∨s)

5. 将条件式中的逻辑或运算符号转换为主析取范式，即将它们转换为逻辑或的形式。

= (p∧¬r∧s)∨(p∧r∧¬s)∨(r∧q)∨(¬r∧¬q∧r)∨(r∧¬q∧¬r)∨(¬r∧s)

因此，主析取范式为 (p∧¬r∧s)∨(p∧r∧¬s)∨(r∧q)∨(¬r∧¬q∧r)∨(r∧¬q∧¬r)∨(¬r∧s)。

Determine whether Player 1 has a winning strategy in the following finite Boolean games, where in both cases \mathrm{\Gamma}_1≔x1,x3 and Γ2≔x2,x4. \psi≔x1∧x3→¬x2∧¬x1→x1∧¬x2→x3∨x4 \psi≔x1∨¬x2∧x2∨x3∧¬x3∨¬x4∧¬x1∨¬x2∨x3∨x4

For the first game, we can construct the following truth table:

x1 | x3 | x2 | x4 | Player 1
---|----|----|----|----------
 0 |  0 |  0 |  0 |    1
 0 |  0 |  0 |  1 |    1
 0 |  0 |  1 |  0 |    0
 0 |  0 |  1 |  1 |    1
 0 |  1 |  0 |  0 |    1
 0 |  1 |  0 |  1 |    1
 0 |  1 |  1 |  0 |    0
 0 |  1 |  1 |  1 |    0
 1 |  0 |  0 |  0 |    0
 1 |  0 |  0 |  1 |    1
 1 |  0 |  1 |  0 |    0
 1 |  0 |  1 |  1 |    1
 1 |  1 |  0 |  0 |    0
 1 |  1 |  0 |  1 |    1
 1 |  1 |  1 |  0 |    1
 1 |  1 |  1 |  1 |    0

In this game, Player 1 wins if and only if the formula $\psi$ is true. From the truth table, we can see that $\psi$ is true for 8 out of the 16 possible assignments of values to the variables. Therefore, Player 1 has a winning strategy in this game.

For the second game, we can also construct a truth table:

x1 | x2 | x3 | x4 | Player 1
---|----|----|----|----------
 0 |  0 |  0 |  0 |    0
 0 |  0 |  0 |  1 |    1
 0 |  0 |  1 |  0 |    0
 0 |  0 |  1 |  1 |    1
 0 |  1 |  0 |  0 |    1
 0 |  1 |  0 |  1 |    1
 0 |  1 |  1 |  0 |    1
 0 |  1 |  1 |  1 |    1
 1 |  0 |  0 |  0 |    1
 1 |  0 |  0 |  1 |    0
 1 |  0 |  1 |  0 |    1
 1 |  0 |  1 |  1 |    0
 1 |  1 |  0 |  0 |    0
 1 |  1 |  0 |  1 |    1
 1 |  1 |  1 |  0 |    0
 1 |  1 |  1 |  1 |    1

In this game, Player 1 wins if and only if the formula $\psi$ is false. From the truth table, we can see that $\psi$ is false for 8 out of the 16 possible assignments of values to the variables. Therefore, Player 1 does not have a winning strategy in this game.