The first (left) part of Law 3 is the same as the end of Step 4, and the second (right) part of Step 3 is the same as the beginning of step 4. So, in Step 4, Law 3 is applied right-to-left.

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- Jul 10th 2011, 05:12 PMemakarovRe: logic related question ...
The first (left) part of Law 3 is the same as the end of Step 4, and the second (right) part of Step 3 is the same as the beginning of step 4. So, in Step 4, Law 3 is applied right-to-left.

- Jul 10th 2011, 05:14 PMiwan1981Re: logic related question ...
Ok so we are allowed to repeat the same step multiple times ...

So now we have:

Step 1: ¬p ∨ q = Law 3

Step 2: q ∨¬p = ...

Step 3: ¬(¬q) ∨ ¬p = Law 2

Step 4: ¬q ⇒ ¬p = Law 3

So now we only need to find out Step 2 ...

Step 2: q ∨¬p = ...

and it looks like that Law 3 is used again here ... is that possible? - Jul 10th 2011, 06:06 PMalexandrosRe: logic related question ...
Iwan, what is the name of the book you are reading in connection with logic

- Jul 11th 2011, 12:47 AMiwan1981Re: logic related question ...Quote:

Iwan, what is the name of the book you are reading in connection with logic

- Jul 11th 2011, 02:38 AMalexandrosRe: logic related question ...
That is your problem.

Unless you get a whole book and start from the beginning and make sure you understand each and every chapter ,then you might have a chance to realise what you are doing and the meaning behind it .

Taking for example the classic book "introduction to logic" by Irving Copi you will find that the problem you are trying to solve is in chapter 9.

How about the 8 chapters before?? - Jul 11th 2011, 04:35 AMiwan1981Re: logic related question ...
I exactly know what you mean ... and I fully agree ...

But the problem is time here :-) ... so I am trying to make the best of it ... and some things are easier to understand then others ... - Jul 11th 2011, 04:51 AMemakarovRe: logic related question ...
One of my instructors used to say, "It is better not to finish anything than not to understand anything."

- Jul 11th 2011, 06:01 AMiwan1981Re: logic related question ...
Back on topic :-)

**So far we still have:**

Step 1: ¬p ∨ q = Law 3

Step 2: q ∨ ¬p = ...

Step 3: ¬(¬q) ∨ ¬p = Law 2

Step 4: ¬q ⇒ ¬p = ...

**The Laws that we can choose from:**

Law 1: (a ∨ b) ⇔ (b ∨ a)

(commutativity of ∨)

Law 2: ¬(¬a) ⇔ a

(law of negation)

Law 3: (a ⇒ b) ⇔ (¬a ∨ b)

(law of implication)

Law 4: (a ⇔ b) ⇔ ((a ⇒ b) ∧ (b ⇒ a))

(law of equivalence)

**Explanation of the steps:**

So with step 1 it was obvious that Step 1 (¬p ∨ q) belonged to Law 3 (a ⇒ b) ⇔ (¬a ∨ b)

Because we see that "¬p" corrosponds with "¬a" and "q" corrosponds with "b"

And with step 3 ( ¬(¬q) ∨ ¬p ) it was also clear that law 2 (a ⇒ b) ⇔ (¬a ∨ b) is the lucky one.

Becasue when we replace the "a" with "¬q" and "b" with "¬p" the result will be (¬q ⇒ ¬p) ⇔ (¬¬q ∨ ¬p)

So the only steps we have left to solve is Step 2 and Step 4...

And the Laws that we have not used yet are Law 1 and Law 4

This is correct until now right? - Jul 11th 2011, 06:11 AMemakarovRe: logic related question ...
Yes, and I also wrote which laws are used in steps 2 and 4 in my previous posts.

- Jul 11th 2011, 06:20 AMiwan1981Re: logic related question ...
- Jul 11th 2011, 06:23 AMemakarovRe: logic related question ...
See post #17, as I also wrote in the private message.

- Jul 11th 2011, 06:28 AMiwan1981Re: logic related question ...
I see now ...

Step 1 = Law 3

Step 2 = Law 1

Step 3 = Law 2

Step 4 = Law 3

You know what my big problem is ...

That is that I was in the understanding that we should use all given laws only 1 time ... and now that Law3 is used twice and Law 4 is not used at all I realise that this is not needed...

We just use the Laws acording to needs ... and this can well be that a Law is used multiple times...

Thanks for your patience with this one :-)