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Math Help - logic related question ...

  1. #16
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    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...
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  2. #17
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    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?
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  3. #18
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    Re: logic related question ...

    Quote Originally Posted by emakarov View Post

    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 ...
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  4. #19
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    Re: logic related question ...

    Quote Originally Posted by emakarov
    Finally, in step 4, you convert a disjunction (q) ∨ p into an implication q ⇒ p. Which law does this?
    Quote Originally Posted by iwan1981 View Post
    I would say that Law 4 does that ...
    Why?
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  5. #20
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    Re: logic related question ...

    Quote Originally Posted by emakarov View Post
    Why?
    because it has a ⇒ sign in t and the other doesn't ...
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  6. #21
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    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.
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  7. #22
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    Re: logic related question ...

    Quote Originally Posted by emakarov View Post
    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 ...
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  8. #23
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    Re: logic related question ...

    What happens when we instantiate a to be q and b to be p in Law 3?
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  9. #24
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    Re: logic related question ...

    Quote Originally Posted by emakarov View Post
    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 ...
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  10. #25
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    Re: logic related question ...

    Replace a with q and b with p.
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  11. #26
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    Re: logic related question ...

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

    Like this?
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  12. #27
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    Re: logic related question ...

    Yes, but there is still one more a to replace with q.
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  13. #28
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    Re: logic related question ...

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

    right?
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  14. #29
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    Re: logic related question ...

    Yes. So, does the left or the right part of this law look like the starting point of step 4?
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  15. #30
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    Re: logic related question ...

    Quote Originally Posted by emakarov View Post
    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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