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 Post subject: Re: proof analysis
PostPosted: Tue, 1 Feb 2011 04:42:03 UTC 
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idi wrote:
Shadow wrote:


Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.

.


idi wrote:

This is not Modus Ponens

The general form of Modus Ponenes is:

[(p=>q) and p] => q

Is the above of that form
??


Shadow wrote:

This IS Modus Ponens. If you cannot recognize it, then you need some more practice with that. P is the statement (A is a field) Q is the statement (A is a ring).


If you put :

P = (A is a field) ...........Q = ( A is a ring).

Can we put then : P=(B is a field) and CONCLUDE ,(B is a ring) is a Q,by using M.Ponenes ??

Yes, A is a dummy object which is part of the statement.

E.g. If X is an NBA player, X is a basketball player.

Now I replace X by "Michael Jordan"

If Michael Jordan is an NBA player, Michael Jordan is a basketball player.

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 Post subject: Re: proof analysis
PostPosted: Tue, 1 Feb 2011 09:07:49 UTC 
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idi wrote:
Shadow wrote:


Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.

.


idi wrote:

This is not Modus Ponens

The general form of Modus Ponenes is:

[(p=>q) and p] => q

Is the above of that form
??


Shadow wrote:

This IS Modus Ponens. If you cannot recognize it, then you need some more practice with that. P is the statement (A is a field) Q is the statement (A is a ring).


If you put :

P = (A is a field) ...........Q = ( A is a ring).

Can we put then : P=(B is a field) and CONCLUDE ,(B is a ring) is a Q,by using M.Ponenes ??


Technically, you need to bind the dummy variables. What you can do is from the statement "for all A, if A is a field, then A is a ring" to "B is a field => B is a ring" (forall-elimination), then from "B is a field => B is a ring" and "B is a field", MP gives "B is a ring" as your desired conclusion. (Or, add enough things to your language and just invoke "B is a field => B is a ring" as an instance of a suitable axiom, then MP as before.)

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Tue, 1 Feb 2011 12:59:29 UTC 
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Then we need the use of two laws :

1) Universal Elimination ,and then

2) Modus Ponens.

in the following way:

for all ,x ( if x is a field ,then x is a ring) [this is a fact in abstract alebra]

Now by using the law of Universal Elimination ,by puting x=A, we have:

If A is a field ,then A is a ring .



Then we have to add but, A is a field , to conlcude A is a ring

The same thing we can say for B

But certainly :

Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


Tell us nothing.

But to come back to our problem ,apart from the law of transitivity ,what other theorems or axioms are involved in the proof??

Also Shadow mentioned the use of M.Tolens .Where is that law involved in the proof??


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 Post subject:
PostPosted: Tue, 1 Feb 2011 13:30:09 UTC 
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idi wrote:
Then we need the use of two laws :

1) Universal Elimination ,and then

2) Modus Ponens.

in the following way:

for all ,x ( if x is a field ,then x is a ring) [this is a fact in abstract alebra]

Now by using the law of Universal Elimination ,by puting x=A, we have:

If A is a field ,then A is a ring .



Then we have to add but, A is a field , to conlcude A is a ring

The same thing we can say for B

But certainly :

Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


Tell us nothing.

But to come back to our problem ,apart from the law of transitivity ,what other theorems or axioms are involved in the proof??

Also Shadow mentioned the use of M.Tolens .Where is that law involved in the proof??


No, that is condensing \forall-Elimination and Modus Ponens into a single line.

You don't have Modus Tollens in the proof you gave for (1) (except: you should prove, or at least examine the proof of, 1>0, where you might encounter it).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Tue, 1 Feb 2011 17:17:31 UTC 
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So ,which are the theorems or axioms and the laws of logic involved in the proof?

So far we have M.ponens M. tolens Universal Elimination and the theorem of transitivity.

Are there any others??

Also i cannot make out yet how these laws are used


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 Post subject: Re: proof analysis
PostPosted: Tue, 1 Feb 2011 18:33:14 UTC 
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idi wrote:
So ,which are the theorems or axioms and the laws of logic involved in the proof?

So far we have M.ponens M. tolens Universal Elimination and the theorem of transitivity.

Are there any others??

Also i cannot make out yet how these laws are used


Again, you should look at your conclusions. You wrote down things like:

idi wrote:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


If you don't know how you concluded them, how do you know they're correct? Learning what techniques you've used involves looking at what you've written, and examining each step to see what you've done. Have you tried this yet?

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 Post subject:
PostPosted: Tue, 1 Feb 2011 21:39:26 UTC 
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Why ,how do you know that my proof is correct??


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