# Thread: logic related question ...

1. ## Re: logic related question ...

Yet, this law does not mention either ∨ or ¬, which are the only connectives (operations) in the formulas (*) and (**) above. Why do you think law 4 applies?
You are right ... and I guess I am wrong with making this statement:

Step 3: ¬(¬q) ∨¬p = Law 4
This is new stuff for me ... and its a bit complicated ...

I think when I look at the laws the one that looks the most the same with Step 3 is Law 2 (Law 2: ¬(¬a) a(law of negation))

But I already used Law 2 in Step 2...

2. ## Re: logic related question ...

I think when I look at the laws the one that looks the most the same with Step 3 is Law 2 (Law 2: ¬(¬a) ⇔ a(law of negation))
Yes.

So far we have

Step 1: ¬p ∨ q = Law 3
Step 2: q ∨¬p = ...
Step 3: ¬(¬q) ∨ ¬p = Law 2
Step 4: ¬q ⇒ ¬p = ...

As you said, the starting point of step 2, which is ¬p ∨ q, looks like a ∨ b from law 1. Finally, in step 4, you convert a disjunction ¬(¬q) ∨ ¬p into an implication ¬q ⇒ ¬p. Which law does this?

3. ## Re: logic related question ...

Originally Posted by emakarov

As you said, the starting point of step 2, which is ¬p ∨ q, looks like a ∨ b from law 1. Finally, in step 4, you convert a disjunction ¬(¬q) ∨ ¬p into an implication ¬q ⇒ ¬p. Which law does this?
I would say that Law 4 does that ...

4. ## Re: logic related question ...

Originally Posted by emakarov
Finally, in step 4, you convert a disjunction ¬(¬q) ∨ ¬p into an implication ¬q ⇒ ¬p. Which law does this?
Originally Posted by iwan1981
I would say that Law 4 does that ...
Why?

5. ## Re: logic related question ...

Originally Posted by emakarov
Why?
because it has a ⇒ sign in t and the other doesn't ...

6. ## Re: logic related question ...

because it has a ⇒ sign in t and the other doesn't ...
Do you mean that one side of Law 4 has a ⇒ sign and the other does not? That's true, but the other side has ⇔, and neither ¬(¬q) ∨ ¬p nor ¬q ⇒ ¬p from Step 4 has it.

7. ## Re: logic related question ...

Originally Posted by emakarov
Do you mean that one side of Law 4 has a ⇒ sign and the other does not? That's true, but the other side has ⇔, and neither ¬(¬q) ∨ ¬p nor ¬q ⇒ ¬p from Step 4 has it.
Ok so I don't know the answer to your first initial question:

As you said, the starting point of step 2, which is ¬p ∨ q, looks like a ∨ b from law 1. Finally, in step 4, you convert a disjunction ¬(¬q) ∨ ¬p into an implication ¬q ⇒ ¬p. Which law does this?
I am totally lost now ... I understand why and how we did the Laws and Steps we already know ... I am just having difficulties in figuring out the last 2 ...

8. ## Re: logic related question ...

What happens when we instantiate a to be ¬q and b to be ¬p in Law 3?

9. ## Re: logic related question ...

Originally Posted by emakarov
What happens when we instantiate a to be ¬q and b to be ¬p in Law 3?
Law 3: (a⇒ b) ⇔ (¬a∨ b) (law of implication)

I have no clue ... really ...

10. ## Re: logic related question ...

Replace a with ¬q and b with ¬p.

11. ## Re: logic related question ...

Originally Posted by emakarov
Replace a with ¬q and b with ¬p.
Law 3: (¬q ⇒ ¬p) ⇔ (¬a∨ ¬p) (law of implication)

Like this?

12. ## Re: logic related question ...

Yes, but there is still one more a to replace with ¬q.

13. ## Re: logic related question ...

Law 3: (¬q ⇒ ¬p) ⇔ (¬¬q ∨ ¬p) (law of implication)

right?

14. ## Re: logic related question ...

Yes. So, does the left or the right part of this law look like the starting point of step 4?

15. ## Re: logic related question ...

Originally Posted by emakarov
Yes. So, does the left or the right part of this law look like the starting point of step 4?
Law 3: (¬q ⇒ ¬p) ⇔ (¬¬q ∨ ¬p) (law of implication)
Step 4: ¬q ⇒ ¬p
Yes the first part of Law 3 is the same ...

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