Proof for the following statement Not Semi-simple

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The discussion centers on the proof that "Not Semi-simple implies degenerate," leading to the conclusion that "non-degenerate implies semi-simple." Participants explore the logical equivalence of these statements through contrapositives, clarifying that if the original statement is true, the contrapositive must also hold. Visual aids, such as geometric representations, are suggested to better understand the relationships between the terms. The conversation emphasizes the importance of logical reasoning and the utility of truth tables for validating these implications. Overall, the discussion highlights the nuances of logical proofs in mathematics.
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I have a proof for the following statement

Not Semi-simple \Rightarrow degenerate.

Does this mean the following is true?

non degenerate \Rightarrow semi-simple
 
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Yes. If your original statement is true, then the second statement, the contrapositive, is also true.
 


I thought so, thanks, I would be happier if I could visualise it in my head though if you know what I mean.

Like 'If it is my car then it is red' then the statement 'if it is not red, then it is not my car' is equivalent and I can easily see that. Just don't quite see it in this case.


Anyways,
Ta
 


It might be because your premise is negated.

Let A mean "semi-simple" and B mean "degenerate"

Your statement is ~A => B

The contrapositive is actually
~B => ~~A

The ~'s cancel to yield ~B => A.

If you were to make up a funny word to mean ~A (maybe "semi-complicated" =-), then it would sound more logical:

"If something is semi-complicated, then it must be degenerate."

"If something isn't degenerate, then it isn't semi-complicated".
 


One way you can visualize it is by using some geometry. Your example: if its my car, then its red. Think of a small circle, in that circle put "my car". Then put that circle in a larger circle, and lable the large circle, "red". Now you can see if you pick your car, you have to pick red. Now if you pick not red, you cannot pick your car.

of course this is by no means rigorous because of the ambiguities of such a set and subset, but its a simple trick that may help you to see what your doing.
 


It won't help with intuition, but just first to convince yourself of the equivalence and its contrapositive you could write out the truth tables of P => Q and ~Q => ~P and compare!
 
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