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Thread: Logic Proofs

  1. #1
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    Exclamation Logic Proofs

    Hey guys these are the problems that I have a hard time doing. I get confused with how the laws work sometimes. If you guys could help me that would be awesome.


    Using the laws of sentential logic and the rules of inference prove the target formulas using the given WFFs:

    1. Given (a → b) and a → c), and a prove (bc).

    2. Given (ab), (ab), and b, prove a.

    3. Given (ab) prove (ab).



    I am also having some issues with this problem as well:

    Give an example of a logical formula involving the atomic statements a, b, c, and d that is always true regardless of the value of these four atomic statements.
    Last edited by marsandfruit; Jul 5th 2017 at 08:47 PM.
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  2. #2
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    Re: Logic Proofs

    Hey marsandfruit.

    Have you every done truth tables before? Have you covered set theory?
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  3. #3
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    Re: Logic Proofs

    From "a" and "a → b", b is true. From "a" and a → c, c is true. Therefore b and c are both true.

    The last asks you to find a tautology. A simple one is "a → a". If you need to find a tautology that includes all of a, b, and c, vary that.
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  4. #4
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    Re: Logic Proofs

    Yes I have. I just have a hard time applying the rules.
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    Re: Logic Proofs

    You said "Using the laws of sentential logic and the rules of inference" so apparently you are not allowed to use a "truth table". Now, what are the "laws of sequential logic" and "rules of inference".
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  6. #6
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    Re: Logic Proofs

    The Laws of Sequential Logic

    Idempotency (p∨p) ⇐⇒ p (p∧p) ⇐⇒ p

    Associativity ((p∨q)∨r) ⇐⇒ (p∨(q∨r)) ((p∧q)∧r) ⇐⇒ (p∧(q∧r))

    Commutativity (p∨q) ⇐⇒ (q∨p) (p∧q) ⇐⇒ (q∧p)

    Distributivity (p∨(q∧r)) ⇐⇒ ((p∨q)∧(p∨r)) (p∧(q∨r)) ⇐⇒ ((p∧q)∨(p∧r))

    Identity (p∨F) ⇐⇒ p (p∧T) ⇐⇒ p

    Domination (p∨T) ⇐⇒ T (p∧F) ⇐⇒ F

    Complement Laws (p∨p) ⇐⇒ T (p∧p) ⇐⇒ F

    Double Complement Law p ⇐⇒ p

    DeMorgan’s Laws (p∨q) ⇐⇒ (p∧q) (p∧q) ⇐⇒ (p∨q)

    Conditional Laws (p → q) ⇐⇒ (p∨q) (p → q) ⇐⇒ (q →p) (p → q) ⇐⇒(p∧q)

    Biconditional Laws (p ↔ q) ⇐⇒ ((p → q)∧(q → p)) (p ↔ q) ⇐⇒ ((p∧q)∨(p∧q)

    Rules of Inference

    Modus Ponens

    It can be summarized as "P implies Q and P is asserted to be true, so therefore Q must be true."


    . It is an application of the general truth that if a statement is true, then so is its contra-positive.
    The first to explicitly describe the argument form modus tollens were the Stoics.[7]
    The inference rule modus tollens validates the inference from {\displaystyle P} P implies {\displaystyle Q} Q and the contradictory of {\displaystyle Q} Q to the contradictory of {\displaystyle P} P.
    The modus tollens rule can be stated formally as:
    {\displaystyle {\frac {P\to Q,\neg Q}{\therefore \neg P}}} {\frac {P\to Q,\neg Q}{\therefore \neg P}}
    where {\displaystyle P\to Q} P\to Q stands for the statement "P implies Q". {\displaystyle \neg Q} \neg Q stands for "it is not the case that Q" (or in brief "not Q"). Then, whenever " {\displaystyle P\to Q} P\to Q" and " {\displaystyle \neg Q} \neg Q" each appear by themselves as a line of a proof, then " {\displaystyle \neg P} \neg P" can validly be placed on a subsequent line.


    and for the last one would this work?

    a → a


    a → b

    a→ c

    b → c
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    Re: Logic Proofs

    Quote Originally Posted by marsandfruit View Post
    Using the laws of sentential logic and the rules of inference prove the target formulas using the given WFFs:

    1. Given (a → b) and a → c), and a prove (bc).

    2. Given (ab), (ab), and b, prove a.

    3. Given (ab) prove (ab).






    1) From $(a \to b) \wedge (a \to c)$ we get $(\neg a \vee b) \wedge (\neg a \vee c)$
    The distrub. gives $\neg a \vee (b \wedge c)$
    But $a$ gives $b\wedge c)$

    2) There is a rule that is not in your list: Disjunctive syllogism. $(p\vee q)\wedge \neg p$ implies $q$.

    3) $a\wedge b$ implies $a$ by simplification ; $a\wedge b$ implies $b$ by simplification.
    Those two give $a\vee b$ by addition.
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