How to proove De Morgan's Law for Logic?? Without using Truth table thanks folks DeMorgan's Law: ~(P^Q)<=>~Pv~Q; ~(PvQ)<=>~P^~Q;
I imagine he wants [tex]-\vee_{i=1}^nk_i=\wedge_{i=1}^n-k_i[/tex] [tex]-\wedge_{i=1}^nk_i=\vee_{i=1}^n-k_i[/tex] (I used - for not; how do you get the proper symbol?)
Well, here's a way to prove the first implication of the first law, ~(p&q) --> ~p v ~q Here the indented lines following an assumption indicate the scope of the assumption: Code (Text): 1. Assume ~(p&q) v (~r&r) 2. (p&q) --> (~r&r) (1, material implication) 3. p --> (q --> (~r&r)) (2, exportation) 4. p --> (~q v (~r&r)) (3, material implication) 5. ~p v (~q v (~r&r)) (4, material implication) 6. (~p v ~q) v (~r&r) (5, association) 7. (~(p&q) v (~r&r)) --> ((~p v ~q) v (~r&r)) (1-6, conditional proof) 8. Assume ~r&r (assumption of line 8 has no scope beyond itself) 9. ~(~r&r) (8, reductio ad absurdum) 10. Assume ~(p&q) 11. ~(p&q) v (~r&r) (10, addition) 12. (~p v ~q) v (~r&r) (7, modus ponens) 13. ~p v ~q (9, 12, disjunctive syllogism) 14. ~(p&q) --> (~p v ~q) (10-13, conditional proof) The second half of the first DeMorgan's law, (~p v ~q) --> ~(p&q), can be proved similarly.
The key to that proof was the use of exportation to get line 7. Both halves of the second DeMorgan's law can also be proved by the same general idea, though it's slightly trickier.
\neg in LaTeX (for negate). Example: [tex] \neg \exists x . P(x) \Leftrightarrow \forall x. \neg P(x) [/tex]
The way I learned this is that is follows from the semantic defintion of truth. Say you you have an interpretation M and a sentence F by definition M|= ~F iff not M|=F M|= (F&G) iff M|= F and M|=G M|= (FvG) iff M|= F or M|=G Thus a consequence of the definition of truth for negation and of two junctions is the fact that (F&G) is true iff ~(~Fv~G) is true, and (FvG) is true iff ~(~F&~G) is true. You have ~(PvQ) iff ~P&~Q, which would be equivalent to ~(FvG) iff ~~(~F&~G), which is the same as saying ~(FvG) iff (~F&~G).
Is there no easier way to prove DeMorgan's theorem without having to use EXPORTATION and DISJUNCTIVE SYLLOGISM rules? Is there a way to prove this Law by just using modus ponens, modus tollens, disjunctive argument, conjunctive argument, simplification, and so on?
What is "and so on"? What rules do you have? Just write them out, provide a link, or tell us the name of the system, if you know it; it shouldn't take but a few minutes.
Code (Text): 1 | ~(P & Q) assumption ...|-------------------- 2 || ~(~P v ~Q) assumption ...||------------------- 3 ||| P assumption ...|||------------------ 4 |||| Q assumption ...||||----------------- 5 |||| P & Q 3, 4 conjunction introduction 6 |||| ~(P & Q) 1, repitition 7 ||| ~Q 4-6 negation introduction 8 ||| ~P v ~Q 7, disjunction introduction 9 ||| ~(~P v ~Q) 2, repitition 10 || ~P 3-9, negation introduction 11 || ~P v ~Q 10, disjunction introduction 12 || ~(~P v ~Q) 2, repitition 13 | ~P v ~Q 2-12 negation elimination Code (Text): 1 | ~P v ~Q assumption ...|----------------------- 2 || ~P assumption ...||---------------------- 3 ||| P & Q assumption ...|||--------------------- 4 ||| P 3, conjunction elimination 5 ||| ~P 2, repitition 6 || ~(P & Q) 3-5, negation introduction 7 || ~Q assumption ...||---------------------- 8 ||| P & Q assumption ...|||--------------------- 9 ||| Q 8, conjunction elimination 10 ||| ~Q 7, repitition 11 || ~(P & Q) 8-10 negation introduction 12 | ~(P & Q) 1, 2-5, 7-11 disjunction elimination
Re: How to proove De Morgan's Law for Logic?? Hi, how to simplify (-p^q)v-(pvq) using the logical equivalence?
Re: How to proove De Morgan's Law for Logic?? How do you prove the following? I understand that it is part of demorgans law.. but how do i get rid of the double negative? -(-A v -B) therefore (A & B)