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Math Help - Implication in logic

  1. #1
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    Implication in logic

    Hi,

    I'm struggling to understand the semantics of nested implications in propositional logic.
    Take the propositions following:
    C: I am eating chocolate
    P: I am drinking Pepsi
    V: I am eating vegetables

    What is the difference between
    (a) (C->P)->V and
    (b) C->(P->V)?

    Here is my understanding so far:
    (a) Given that I am drinking Pepsi, as a consequence of eating chocolate, I am eating vegetables.
    (b) I am only eating vegetables if I have drunk Pepsi, which in turn was preceded by eating chocolate.

    How can I translate this into English, in sensible terms?

    Thank you!
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  2. #2
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    Re: Implication in logic

    Read it off the definition of A->B:

    C P (C->P) V (C->P)->V
    T T T T T
    T F F F T
    F T T T T
    F F T F F

    It's a table.Sorry, I can't get it to line up. Multiple spaces don't show.

    An explanation of the truth table is given in:

    Truth Tables, Tautologies, and Logical Equivalence
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  3. #3
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    Re: Implication in logic

    Quote Originally Posted by aprilrocks92 View Post
    What is the difference between
    (a) (C->P)->V and
    (b) C->(P->V)?

    Here is my understanding so far:
    (a) Given that I am drinking Pepsi, as a consequence of eating chocolate, I am eating vegetables.
    Pretty much. I would say, "If (my) eating chocolate implies drinking Pepsi, then I am eating vegetables".

    Quote Originally Posted by aprilrocks92 View Post
    (b) I am only eating vegetables if I have drunk Pepsi, which in turn was preceded by eating chocolate.
    First, "only if" is the converse of "if", so "I am only eating vegetables if..." corresponds to V -> ... . Second, there is no past tense in propositional logic, so you can't say "was preceded by". This can be rendered in (awkward) English as "If I am eating chocolate, then if I am drinking Pepsi, then I am eating vegetables", which is eqivalent to "If I am eating chocolate and drinking Pepsi, then I am eating vegetables".

    You see that C -> (P -> V) is equivalent to C /\ P -> V. In both cases V holds if both C and P hold. This equivalence can also be checked using a truth table. Thus, implication nesting on the right does not create complexity: the two premises can be joined by conjunction to remove the nested implication. In contrast, (C -> P) -> V is a real nested implication than creates complexity. I think it is much harder to comprehend ((A -> B) -> C) -> D than A -> (B -> (C -> D)).

    Using the equivalences C -> P = ~C \/ P and (A \/ B) -> C = (A -> C) /\ (B -> C) or using a truth table, it is easy to show that (C -> P) -> V = (~C -> V) /\ (P -> V). But the first equivalence above does not hold in some areas of math (for example, in topos theory). In these areas there are proofs that proceed by induction on the implication nesting on the left as the complexity measure of formulas.

    In sum, C -> (P -> V) is almost the same as C /\ P -> V, but (C -> P) -> V is something very different. Note that C -> P -> V without parentheses is usually parsed as C -> (P -> V).
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  4. #4
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    Re: Implication in logic

    It's easier to read it off the truth table for "->", which defines implication. See my previous reference.

    My apologies for the lousy table. I'll try to improve it by first doing it in Word and see what happens.

    EDIT: If you want to analyze it syntactically, you have to first define "implication."
    Last edited by Hartlw; September 9th 2013 at 07:52 AM.
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  5. #5
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    Re: Implication in logic

    Hartlw, for tables you can use [code]...[/code] tags, which preserve alignment. Also, there are BB tags for tables similar to HTML tags, but in square brackets.
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  6. #6
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    Re: Implication in logic

    Quote Originally Posted by aprilrocks92 View Post
    Hi,

    I'm struggling to understand the semantics of nested implications in propositional logic.
    Take the propositions following:
    C: I am eating chocolate
    P: I am drinking Pepsi
    V: I am eating vegetables

    What is the difference between
    (a) (C->P)->V and
    (b) C->(P->V)?

    Here is my understanding so far:
    (a) Given that I am drinking Pepsi, as a consequence of eating chocolate, I am eating vegetables.
    (b) I am only eating vegetables if I have drunk Pepsi, which in turn was preceded by eating chocolate.

    How can I translate this into English, in sensible terms?

    Thank you!
    The simplest definition of implication, A->B, sufficient for this problem, is that B is true if A is true, which is either a definition, a statement of fact, or nonsense.
    1) a-b positive -> a>b, definition
    2) it’s less than 32dedF-> pure water freezes, fact
    3) its Tuesday->it’s cold, nonsense

    (A->B)->C says: If AimpliesB is true, then C is true
    A->(B->C) says: If A is true, then BimpliesC is true.

    (C->P)->V can be expressed as an english sentence but it won’t make sense (be understandable) unless you are an individual who always drinks Pepsi when he eats chocolate and you have a doctors certificate that says you eat vegetables because you always drink Pepsi when you eat Chocolate. Simply eating vegetables doesn’t cut it (there has to be causality).

    So the answer to your question is you can create an english sentence in this case but it won’t make sense.
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    Re: Implication in logic

    Quote Originally Posted by Hartlw View Post
    C P (C->P) V (C->P)->V
    T T T T T
    T F F F T
    F T T T T
    F F T F F
    Personally I dislike the "code" box. It's annoying trying to get everything to line up. You could try this:
    \begin{array}{c|c|c|c|c} C & P & C \to P & V & (C \to P) \to V \\ \hline T & T & T & T & T \\ \hline T & F & F & F & T \\ \hline F & T & T & T & T \\ \hline F & F & T & F & F \end{array}

    Which gives:
    \begin{array}{c|c|c|c|c} C & P & C \to P & V & (C \to P) \to V \\ \hline T & T & T & T & T \\ \hline T & F & F & F & T \\ \hline F & T & T & T & T \\ \hline F & F & T & F & F \end{array}

    -Dan
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  8. #8
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    Re: Implication in logic

    After all this while, why this answer right after my last Post; especially after my last Post made use of tables irrelevant and I would just as soon not have the one you quote emphasized. Why not put your response in my "Tables" post, where I asked about tables? At best, it's a matter of questionable style.
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    Re: Implication in logic

    The truth table for (A->B)->C in the previous post is wrong, which is why it annoyed me. The correct truth table is given below.

    A implies B is a sentence (subject-predicate-object) no matter what A and B are as long as they are noun equivalents.

    it can make sense depending on A and B and the defintiion of “implies.” An example is given in post #6

    by definition, it is a “logical” sentence, A->B, if A, B and the sentence can take on one of two values: (T,F); (0,1) ; (*,$) ; etc. In a sense it is a “function” (not 1-1) of two variables:

    A->B
    A.B.A->B
    T.T.T
    T.F.F
    F.T.T
    F.F.T

    The values of the variables and “function” are motivated by the conventional definition of “implies.”

    An analogy is a real and abstract painting.
    A painting consists of shapes and colors on a flat surface.
    “Therefore” any collection of shapes and colors on a flat surface is a painting. Depending on the shapes and colors, the painting may or may not be real (logical).

    Personally I do not believe that abstract painting, sculpture, or architecture is the foundation of, or necessary for, art. Michelangelo got along fine without it.

    Truth table for (A->B)->C. Let F=(A->B) and G=F->C (so I don’t have to do a table).
    A.B.C.F.G
    T.T.T.T.T
    T.T.F.T.F
    T.F.T.F.T
    T.F.F.F.T
    F.T.T.T.T
    F.T,F.F.T
    F.F.T.T.T
    F.F.F.T.F
    The truth table for A->(B->C) gives the same result so that
    (A->B)->C = A->(B->C)

    Edit: Copied table wrong. Corrected.
    Last edited by Hartlw; September 14th 2013 at 07:10 AM.
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  10. #10
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    Re: Implication in logic

    Heck. Couldn't catch the "edit-" not much time. Oh well:

    EDIT If F(A,B)=A->B, then F(F(A,B),C) is (A->B)->C

    Had a little trouble with this at first. As a rough analogy, suppose z=f(x,y) and x = g(u,v). Then f(g(u,v),y) is a function of three variables. But suppose g=f so that z=f(f(u,v),y), which somehow looks circular. But if you use a real example, it seems to be OK.

    If f(x,y)=x^2+y^2 and f(u,v)=u^2+v^2.
    Then f(f(u,v),y) = (u^2+v^2)^2+y^2.
    Last edited by Hartlw; September 14th 2013 at 09:01 AM.
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