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Math Help - Logic Question

  1. #1
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    Logic Question

    I'm reading a book about discrete math, and am learning about the rules of inference. Here's the problem:

    p = "You have access to the network."
    q = "You can change your grade."

    Given:
    i.) p--> q is false
    ii.) p is true.

    The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.

    The book says these p and q are propositions, NOT predicates. Am I missing something here, or is this a mistake by the author? My only guess is that the author is miswrote that p and q are propositions, and in fact is treating p and q as predicates where

    p = x has access to the network
    q = x can change his grade

    In which case "you" are in the domain of discourse, and there exists x1 in the domain of discourse such that p(x1)^not(q(x1)) is true. In this case, it would make sense to state that the conclusion is undecidable.
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  2. #2
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    Quote Originally Posted by mbalaban View Post
    I'm reading a book about discrete math, and am learning about the rules of inference. Here's the problem:

    p = "You have access to the network."
    q = "You can change your grade."

    Given:
    i.) p--> q is false
    ii.) p is true.

    The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.

    The book says these p and q are propositions, NOT predicates. Am I missing something here, or is this a mistake by the author? My only guess is that the author is miswrote that p and q are propositions, and in fact is treating p and q as predicates where

    p = x has access to the network
    q = x can change his grade

    In which case "you" are in the domain of discourse, and there exists x1 in the domain of discourse such that p(x1)^not(q(x1)) is true. In this case, it would make sense to state that the conclusion is undecidable.
    [Edit: like this post is wrong. See below.]

    My understanding is that when you take a predicate and specify a subject, it becomes a proposition. Here, I think it was intended that "you" is not the general term for "somebody" but rather the particular person, you (the reader).

    I would write your (i) as

    i) (p -> q) is false

    just to be extra safe that nobody interprets it as

    i) (p -> (q is false)) is true

    I can give an illustration of why the book's conclusion is right: Say every Monday when you can access the network, you can change your grade, but only on Monday. So accessing the network is not sufficient for changing your grade, meaning (p -> q) is false. But suppose you have access to the network on all days, meaning p is true. Now given all these facts, we cannot conclude whether q is true, because we don't know what day of the week it is.
    Last edited by undefined; July 20th 2010 at 01:21 PM.
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  3. #3
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    Perhaps another way of thinking about it is this: the fact that the implication is false doesn't imply that there is only one possible truth assignment to the constituent propositions. It only means that p true and q false is possible.

    [EDIT] See posts 4 and 5 below for a correction to this post.
    Last edited by Ackbeet; July 20th 2010 at 12:14 PM. Reason: Incorrect reasoning here.
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  4. #4
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    Given:
    i.) p--> q is false
    ii.) p is true.

    The book says that it cannot be concluded whether or not you can change your grade. It seems to me that it can be it can be concluded that q is false. We know that p is true and p-->q is false; it can be concluded that q is false because p-->q is false only when p is true and q is false.
    I absolutely agree with you; I don't understand why the author claims what he does.

    Say every Monday when you can access the network, you can change your grade, but only on Monday. So accessing the network is not sufficient for changing your grade, meaning (p -> q) is false. But suppose you have access to the network on all days, meaning p is true. Now given all these facts, we cannot conclude whether q is true, because we don't know what day of the week it is.
    I believe that whether this example applies depends on what the topic the book is teaching. If the topic is the context of reasoning, e.g., adding new, previously hidden, propositions that may change the truth value of the old ones, then yes, the example is relevant. However, if the topic teaches regular propositional logic and regular rules of inference, it is probably a stretch.

    Perhaps another way of thinking about it is this: the fact that the implication is false doesn't imply that there is only one possible truth assignment to the constituent propositions. It only means that p true and q false is possible.
    Again, if the topic is modal logic, possibility and necessity, then OK. It seemed to me that the OP meant regular introductory discreet math textbook.
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  5. #5
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    On second thought, after reading emakarov's post, I'd have to say that if you look at the truth tables involved, you are forced to the conclusion that q is false. It's the only row of the table that works. p is true, q is false, and p -> q is false. So please disregard Post #3.
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  6. #6
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    [Edit: like this post is wrong. See below.]

    Hmm, here is a more "mathy" example.

    Let x be a real number.

    Then the statement: x^2 = 4 \implies x=2 is false, because we could also have  x = -2.

    Given:

    (i) x^2 = 4 \implies x=2 is false.
    (ii) x^2 = 4 is true.

    Surely we can't conclude that \display x \ne 2 ? Am I missing something?

    If you have a problem with "Let x be a real number" then put it this way.

    a = "x is a real number"
    b = " x^2 = 4"
    p = a \land b
    q = "x = 2"
    Last edited by undefined; July 20th 2010 at 01:22 PM.
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  7. #7
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    I think you can conclude that x\not=2. You can think of the implication p\to q as \neg p\vee q, right? Negate this, now, to get \neg(\neg p\vee q)\equiv p\land\neg q.

    Therefore, technically, you don't even need the second premiss in the OP in order to conclude that p is true and q is false.
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I think you can conclude that x\not=2. You can think of the implication p\to q as \neg p\vee q, right? Negate this, now, to get \neg(\neg p\vee q)\equiv p\land\neg q.

    Therefore, technically, you don't even need the second premiss in the OP in order to conclude that p is true and q is false.
    Thanks, after looking at the truth tables, I'm given to believe that you are right, nevertheless I'm puzzled as to why my examples don't apply. Maybe something to do with an indeterminate?

    Sorry for misinformation ..

    Edit: I now definitely believe the discrepancy is due to indeterminates; for example there's no way to fill in values for x and still have it appear the way it did in post #6:

    (i) 2^2 = 4 \implies -2=2 is false.
    (ii) 2^2 = 4 is true.

    We can therefore conclude that (-2=2) is false, and like Ackbeet said, we don't even need (ii) to arrive at this conclusion. In fact, the truth of (i) implies the truth of (ii).

    So, in the example with days of the week (post #2), I got rid of one indeterminate ("To what person does 'you' refer to?") and replaced it with another indeterminate ("What day of the week is it?"). And in the example with x^2 = 4, x is an indeterminate. So I confused propositional logic with first-order logic (of predicate logic). Sorry again.
    Last edited by undefined; July 20th 2010 at 01:25 PM.
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