I also know that the negative proposition and the original proposition are equivalent and skillful. For example, if P deducts Q, in addition to what I said above, there is a necessary condition of non-P and non-Q, and a sufficient condition of non-Q and non-P. You can do the problem.
2.P is cheap, Q is bad, non-P is not cheap, and non-Q is good. If P is a necessary and sufficient condition of Q, then non-Q is a necessary and sufficient condition of non-P. "Cheap goods are not good" and "good goods are not cheap" are mutually negative propositions and are equivalent. But "neither cheap nor good goods" and "good goods are cheap" are equivalent and mutually negative propositions, which are not mentioned in the title.
3.P is "a≠ 1 or b≠2" and q is "a+b≠3", which may be difficult to judge, but it is not "a =1and b=2" (non-propositional or interchangeable) or "a". Obviously, non-P leads to non-Q, and non-Q cannot lead to non-P. Non-Q is a necessary and sufficient condition of non-P, and P is a necessary and sufficient condition of Q.
Direct positive consideration: for example, a=0 and b=3 conform to the P proposition, and a+b=3 is obtained, that is, P cannot deduce Q. ..
On the other hand, if a+b is not equal to 3, it is absolutely impossible for a= 1 and b=2, so a≠ 1 or b≠2 is correct. You need to understand here, because there is no counterexample, don't get stuck. For example, a= 1, b=3, but b≠2, no matter what example is given, the former one can still be deduced later.
This is not a logical problem, but a conceptual problem. We can't take it for granted when it comes to the signing of prescriptions.
For example, a=- 1, b=-2, a > b and a 2.
a=-2,b= 1,a^2>b^2,ab^2,a^2-b^2>； 0, (a+b)(a-b)>0, and the sign of a+b is unknown, so it is impossible to judge whether a-b is positive or negative.
Reverse a-b > 0, both sides are multiplied by (a+b) at the same time, because we don't know the sign of a+b, so we can't continue to deduce.