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Help with logic exercises

  1. Feb 6, 2012 #1
    I'm having some troubles about some exercises regarding my first year logic course.

    1) Deriving (P -> Q) \/ (Q -> R) and showing that this statement is a tautology without using truth tables or venn diagrams.

    So far I have no clue on how to start this question. From what I remember in class, a tautology is a statement that is true for all domains and predicates.

    2) Deriving P <-> P /\ (P \/ R) without using truth tables or venn diagrams and without using the absorption method.

    P <-> P /\ (P \/ R)
    P /\ (Q \/ ~Q) <-> (P /\ P) \/ (P /\ Q) identity (left) distributive (right)
    <-> P \/ (P /\ Q) idempotency (right)

    Thats as close as I can get, but I'm still stuck. I was trying to get both sides similar.

    3) Four classmates (William, Xavier, Youssef and Zachary) were suspected of cheating on their homework. At separate meetings with their instructor, they said the following:

    William: If Xavier cheated, so did Zachary.
    Xavier: William cheated, but Zachary did not.
    Youssef: I did not cheat, but at least one of William or Zachary did.
    Zachary: If William did not cheat, then Youssef did.

    Let W represent the statement: William cheated"
    X represent the statement: Xavier cheated"
    Y represent the statement:Youssef cheated",
    Z represent the statement: Zachary cheated"

    and note that each student either cheated or they did not cheat.

    (a) If each student is telling the truth, which student(s) cheated?
    (b) If the students who cheated did not tell the truth at the meeting, and the students who did not cheat
    did tell the truth, which student(s) cheated?
    To answer the questions, you may nd it helpful to express each students' statement using a logical expression.
    Do not use truth tables
  2. jcsd
  3. Feb 7, 2012 #2
    So, if you're not allowed to use truth tables explicitly, what tools can you use? Did your course define any axiom system for propositional calculus?
  4. Feb 7, 2012 #3


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    Can you use the equivalence P-->Q == ~P \/Q ?
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