• Nicholas Ferreira
    78
    Hello. I spent some time trying to prove the validity of De Morgan's theorems in propositional logic by direct proof and reductio ad absurdum, but I couldn't do the demonstration without assuming that they are already valid, which would be a circular reasoning. Does anyone know how to do this? I know it's easy to do it using truth tables, but I'd like to do this based on other rules of inference. Thank you all!
    1'st De Morgan Theorem: ¬(P∨Q)↔(¬P∧¬Q)
    2'st De Morgan Theorem: ¬(P∧Q)↔(¬P∨¬Q)

  • fdrakeAccepted Answer
    6.7k
    For 1 maybe something like:

    1. ~(PvQ)
    2. Assume P.
    3. PvQ (2, disjunction introduction)
    4. ~P (1,2,3, contradiction).

    Repeat for Q, though it might just give you the => implication. In case that's true, something like:

    1: ~P and ~Q
    2. Assume P v Q
    3. Assume P
    4. ~P from contradiction
    5. Assume Q
    6. ~Q from contradiction
    7. ~(P v Q)

    would probably work for the <=.

    For 2.
    1. ~P or ~Q
    2. Assume P & Q
    3. Assume P
    4. ~Q (disjunctive syllogism, 3,1)
    5. Assume Q
    6. ~P (disjunctive syllogism, 4,1)
    7. ~(P & Q) (2, contradiction)

    Same qualifying remark for the reverse implication - this might not be a reversible proof.

    I don't know if these strategies are actually valid in the logic you're working in, but perhaps they might inspire you to find a strategy which works.
  • Nicholas Ferreira
    78
    Thank you! I totally forgot about making assumptions in the argument to build the proof. I took the way you made the first proof and modified a little to fit in the reductio ad absurdum proof. The same with the second. My goal was to proof that the consequent follows the antecedent by reductio ad absurdum, and it's done. Here the way i did:

    For 1:
    1.   ¬(PvQ)
           ∴ (¬P∧¬Q)
    2.   assume: ¬(¬P∧¬Q)
    3.      | assume: P
    4.      |    | PvQ (3, add.)
    5.      | ∴ ¬P (3; 1 contradicts 4)
    6.      | assume: Q
    7.      |    | QvP
    8.      | ∴ ¬Q (6; 1 contradicts 7)
    9.      | ¬P∧¬Q (5,8, conj.)
    10. ∴ ¬P∧¬Q (2; 2 contradicts 9)
    

    For 2:
    1) ¬(P∧Q)
         ∴ ¬P∨¬Q
    2) assume: ¬(¬P∨¬Q)
    3)     | assume: ¬P
    4)     |     | ¬P∨¬Q (add.)
    5)     | ∴ P (de 3; 2 contradicts 4)
    6)     | assume: ¬Q
    7)     |     | ¬Q∨¬P
    8)     | ∴ Q (de 6; 2 contradicts 7)
    9)     | ∴ P∧Q (5,8, conj.)
    10) ∴ ¬P∨¬Q (3; 1 contradicts 9)
    
  • TheMadFool
    13.8k
    Hi Nicholas. You're learning logic. I have a question for you. You used <=> in your OP as in ~(p v q) <=> ~p & ~q

    I've seen another way to express this with :: as in ~(p v q) :: ~p & ~q

    What's the difference between double implication (<=>) and logical equivalence (::)?
  • Nicholas Ferreira
    78
    Hello. Well, in the books I read it didn't talk about this difference, and it seems, to me, that they are logically the same thing. But, intuitively, I would say that double implication means what the name states, i.e., if P↔Q, then we know that P⊃Q and Q⊃P.
    On the other hand, if two sentences P and Q are logically equivalent, then they express the same logical truth (and therefore have the same truth value), and one of them can be substituted by the other without problem. I don't know if it is right, but the difference seems, for me, to be a semantical one. Do you know the difference, or some book or paper that explains more about this? Thanks for the comment!
  • fdrake
    6.7k


    Propositional and first order predicate logic both have the deduction metatheorem. They also have the property of completeness, so that semantic entailment (truth tables or tableaux) and syntactic entailment (proofs in the logics using the inference rules) are equivalent.
  • TheMadFool
    13.8k


    A double implication like P <-> Q is true when both P and Q have the same truth values in a truth table.

    Two propositions P and Q are logically equivalent iff they have the same truth values in a truth table.
  • MindForged
    731
    they're they same thing.
  • fdrake
    6.7k


    They're the same thing, since propositional logic is complete and has the deduction theorem.

    You just have to be careful with what those iffs work like. It's true in propositional logic that If P=>Q this yields P |- Q and P |=Q, but the 'this yields' isn't the same thing as material implication. Saying P |- Q (syntactic derivation) <=> P |= Q (semantic derivation) can be flat out wrong since the <=> biconditional is syntactic. When you're dealing with languages and metalanguages at the same time it pays to be careful with your symbols.
  • Nicholas Ferreira
    78
    Thank you for answering! I'll read the links you sent. Do you have any recomendation for me to go deeper on propositional logic (and also, what does "first order" logic means? I read a little about it but I think I didn't get the idea)? I read some books of introduction to logic (H. Gensler, I. M. Copi and C. Mortari), but, as the title say, they are just introductions, nothing much deep. I'm interested in those "meta-logic" things, but I don't know exactly what to search for.
  • fdrake
    6.7k


    First order logic is logic with quantification over variables. It's an expansion of propositional (or zeroth order) logic. It's been a while since I studied logic, but I can recommend the book 'Logic with Trees', it goes through propositional and first order logic thoroughly and includes some stuff on soundness/completeness/other things in meta-logic. Lots of exercises too.
  • Nicholas Ferreira
    78
    Nice, thanks for the explanation and the recomendation. I'm downloading the book and I'll read it. I read the summary and it seems to be very good. Philosophy of logic and metalogic are very interesting for me. Hope it helps me. :D
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