what is the negation" all cabbages are green

- anonymous

what is the negation" all cabbages are green

- schrodinger

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- anonymous

be more specific

- anonymous

suppose that "if z is a w-group then z is solvable" is a true statement, you also know that the "z is solvable" is true. can you deduce that "z is a w-group" is true?

- anonymous

i wish i was good a math

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## More answers

- anonymous

no because for the statement to be true z is a w-group could be true or false

- lgbasallote

unless there's more to that question...don't you just put a "not" before the adjective?

- anonymous

igbasallote what are you talking about?

- anonymous

in a "if p then q" statement. if "if p" is false the whole statement is true regardless of then p. If they are both true then it is also true. It can only be false if "if p" is true and "then q" is flase.

- anonymous

I can give you an example to clarify if you want

- anonymous

I was answering ur second question by the way

- anonymous

@lgbasollote was answering your first.

- anonymous

and @lgbasallote you were correct

- anonymous

I don't get what lgbasallote was saying

- anonymous

actually let me look at me logic notes to make sure.

- anonymous

The correct answer to your first question is All cabages are not green

- anonymous

Like the car is red. The negation is the car is not red

- anonymous

why isn't it not all cabbages are green

- anonymous

because that implies that some are green

- anonymous

we want to to do a complete opposite

- anonymous

its confusing I know

- anonymous

so u wan't to tell that none of them are green right?

- anonymous

correct

- anonymous

Did you understand what I was saying to your second question

- anonymous

im trying to figure it out now

- anonymous

What class is this? just curious

- anonymous

advanced mathematics

- anonymous

whole class based on proofs basically

- anonymous

induction proofing, contradictions, and now we are at sets theories

- anonymous

I took a class called mathematical logic which involved a bunch of this stuff. Kinda interesting kinda hell

- anonymous

I hate sets

- anonymous

Only class I have ever had to drop cuz i was failing. Sets, relations, and functions.

- anonymous

Let me know if you don't understand my explanation about your second question I can provide an example that would make it clearer. I'm going to move on to other questions

- anonymous

thats pretty much what i am in now, and let me tell you that I have no idea how im going to pass that class

- anonymous

Logic was kinda fun. Sets is hell. I would master definitions. completely understand what a subset is and what real numbers and complex numbers. just master definitions because you can use those in ur proofs. I wish I had done that.

- anonymous

im trying to figure out that example and don't completely get it
So P=>Q so if z is a w-group then z is solvable IS TRUE, then "Z is solvable is also true"
why s s "Z is a w-group false?

- anonymous

The negation of "all cabbages are green" is "there exists a cabbage which is not green". All is a quantifier.

- anonymous

isn't the negation "all cabbages are not green?"

- anonymous

So @DanielxAK we just had to show that for all to be false one is true?

- anonymous

lets use p and q for your question all the words are confusing me lol. So the question is essentially... If, if p then q is true, then q is true" So its a nested statement?

- anonymous

P-->Q T Premise
Q T Prove

- anonymous

I'm not sure I understand what you mean by that. If it was phrased, cabbage is green. Then, the negation would be cabbage is not green. But, you have "all cabbages are green". So, the negation would be "there exists a cabbage which isn't green". The negation of all is one.
This might be able to explain it better than I can:
http://www.math.cornell.edu/~hubbard/negation.pdf

- anonymous

Good catch Daniels

- anonymous

Your second question @math_proof, I think it is a correct statement. Because "if p" is true, for the statement to be true (which it is by a premise) "then q" must be true. and "if p" is false. Then "then q" is assumed true because it can't be proven otherwise.

- anonymous

You got me inerested now. I dug out my logic notes.
Law of Equivalence for Implication and Disjunction (LEID) states IF p-->q is true, THEN not p or q is true

- anonymous

still confused omg

- anonymous

So we are Given that P --> Q is true correct? and we have to prove that Q is true? am I understanding right?

- anonymous

we know that Q is true

- anonymous

so P can be either true of false

- anonymous

so P=>Q, z is a w-group then z is solvable" is a true statement, so P->Q is true and we know Q is true because "Z is solvable" so P can be either True of False? thats what i'm thinking

- anonymous

P->Q T Premise
Q T Premise
P T Prove
Ok I'm on the right page

- anonymous

So I would say the answer is ~P V P

- anonymous

thats based on truth table on logical implication

- anonymous

I agree

- anonymous

so we can't deduce if P is true

- anonymous

I have this cheat sheet that my teacher made us with all the laws of logic that can be used in a proof. If you'd like it i'll email it to you. Just private message me your email. If you are not comfortable with that I guess I could post it on this thread, but I've got warned for stuff like that

- anonymous

I guess we can't for certain

- anonymous

omg thanks so much for that, I actually get it now

- anonymous

No problem. When you jump into sets I will be no help. Like I said earlier for sets proofs I would master definitions because my teacher used a lot of "By def'n..." in her proofs

- anonymous

i never heard of this website, but now that i discovered it is so useful and helpful and for free

- anonymous

how do you give a medal?

- anonymous

oo oki got it

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