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Determine if the logical equivalent is valid for all predicates P and Q.

  1. Mar 5, 2017 #1

    s3a

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    1. The problem statement, all variables and given/known data

    There are four (sub)problems in the attached PDF file, but I don’t see the common pattern that the solutions to these kinds of problems have; see “3. The attempt at a solution”.

    2. Relevant equations
    Logical equivalence

    3. The attempt at a solution
    I am trying to find the common pattern that the solutions to these kinds of problems have, and for example, for 8a, the solutions PDF is showing that (i) if the LHS is true, then the RHS is true as well as (ii) if the RHS is true, then the LHS is also true, whereas, for 8d, the solutions PDF is showing that (i) if the RHS is true, then the LHS is true and (ii) if the RHS is false, then the LHS is also false.

    Basically, why aren’t both 8a and 8d showing that (i) if the LHS is true, then the RHS is true as well as (ii) if the RHS is true, then the LHS is also true (the "technique", for the lack of a better word, used in 8a), or alternatively, that (i) if the RHS is true, then the LHS is true and (ii) if the RHS is false, then the LHS is also false. (=the technique used in 8d)?

    What’s the logic behind the using of different techniques?

    I feel like I can follow the solutions for 8a,8c and 8d, but I cannot replicate them (without memorizing), because I don’t understand the common pattern they share, so any help in fully understanding these problems would be GREATLY appreciated!

    P.S.
    The (sub)problem 8b seems to make sense to me, but it seems to not share the type of techniques used for 8a, 8c and 8d (probably since it’s just about finding values to disprove the stated logical equivalence, which is different from showing that a stated logical equivalence always holds true).

    P.P.S.
    If anything I typed is unclear, please let me know.
     

    Attached Files:

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  2. jcsd
  3. Mar 6, 2017 #2

    haruspex

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    It tends to be driven by the outermost operator.
    If it is ∀ (as in d, LHS) then it is usually easier to proceed by reductio ad absurdum, i.e. suppose it is false. That allows you to say "so ∃ ...".
    Conversely, if it is ∃ (as in a, LHS), assuming it is true allows you to proceed that way.
    If it is ∨ (a, RHS and d, RHS) then you can break it into two cases.
    If it is ∧ then you can follow both paths at once.
     
  4. Apr 28, 2017 #3

    s3a

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    Thanks for your input, and sorry for my very-late reply!

    Okay, so for parts c and d, we want to show that (i) RHS being true implies that LHS is true and (ii) the LHS being true implies that RHS is true, but we show (ii) by showing its contrapositive, right?

    Assuming I'm correct (so please confirm), what I just said above clarifies my confusion a lot (in addition to what you typed in your reply).
     
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