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    Hey, I've ben attempting some D2 (a bit unsucessfully, I say ashamedly), and I stumbled across this question which I can make neither head nor tail of (even with the answer)

    the question is (for those of you with books, MEI D2/C Ex. 4E Q.7)

    Express the proposition p \Rightarrow q using only p, \ q, ~ and \land

    I have literally no idea how to go about this...

    Apparently the answer is ~  (p \land ~ q ) . Can someone please explain to me how this is equivalent?

    EDIT: I've just drawn up a truth table, and I can see why this is equivalent (well, I can see that it IS equivalent, but I don't understand WHY this is the case?). And how would I get from the first one to the second one?
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    Using my Logic taught in Computing.

    From the answer:

    If p is false, then the statement is always true (False AND something is always false, and the negation of False is true - thus the statement is true).
    If p is true, then you have two cases:
    q is true, in which case ~q is false and the entire thing evaluates to true.
    q is false, in which case the entire thing is false.

    This should agree with the definition of 'implies' (that, False => Anything is always true, and True => B is B).

    I hope this helps.
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    (Original post by foolsihboy)
    Hey, I've ben attempting some D2 (a bit unsucessfully, I say ashamedly), and I stumbled across this question which I can make neither head nor tail of (even with the answer)

    the question is (for those of you with books, MEI D2/C Ex. 4E Q.7)

    Express the proposition p \Rightarrow q using only p, \ q, ~ and \land

    I have literally no idea how to go about this...

    Apparently the answer is ~  (p \land ~ q ) . Can someone please explain to me how this is equivalent?

    EDIT: I've just drawn up a truth table, and I can see why this is equivalent (well, I can see that it IS equivalent, but I don't understand WHY this is the case?). And how would I get from the first one to the second one?
    If you consider, when is p \Rightarrow q ever false? There is only one combination of truth values for p,q where this is the case.

    Then you just negate it to find when p \Rightarrow q is true.
 
 
 
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