Discrete Mathematics implies an equivalent preference as a false proposition for true understanding.

Equivalent formula: p-> q <=> ~ Pvq: How do I understand p-> q as a true proposition when P is false?

The implication p-> q indicates that if p then Q, obviously: If P is true, Q is true, P → Q is a true proposition, and when P is true, when Q is a false proposition, P → Q is a false proposition. For example, Zhang San said, "If it doesn't rain tomorrow (P), then he will go to your house to play (Q)." If it doesn't rain the next day, he will go to your house, he told the truth (P → Q is true). If it doesn't rain, but he didn't go to your house, apparently he lied (P → Q is false at this time ).

However, when P is false, whether Q is a true or false proposition at this time, P → Q's true or false seem to be unable to be judged, and if it rains the next day, no matter how often John doesn't go to your house, zhang San's words cannot be judged to be true, but he does not say anything. In this sense, Zhang San's words are still true, which is called a "good faith assumption". Therefore, we stipulate that, false P indicates that P → Q is true. For example, the proposition "If 2 + 3 = 4, then the sun goes out of the East", "If 2 + 3 = 4, then the sun came out from the West. "They all think it is a true proposition. consider an example in mathematics." If x> 2, x + 1 is ≥3 ", obviously, this proposition is true for any real number X, but when X is 3, 2, and 1 respectively, the preceding proposition is "If 3> 2, 3 + 1 ≥ 3 ", "If 2> 2, 2 + 1 ≥ 3", "If 1> 2, 1 + 1 ≥ 3", it can be seen that when and only when P is true, when q is false, P → Q is false, and the rest are true.

 

Original article address:Http://www.itlogger.com/program/1451.html