May 12th, 2016, 04:16 PM  #1 
Newbie Joined: May 2016 From: London Posts: 7 Thanks: 0  Laws Of Logic
Use the laws of logic to prove that ¬(p∧ q) ∨ (¬p∧ q)== ¬p V ¬q Any Help would really be appreciated. Thanks 
May 12th, 2016, 05:26 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 688 Thanks: 243 
What laws of logic do you know? Do you know De Morgan's Laws?

May 12th, 2016, 05:37 PM  #3 
Newbie Joined: May 2016 From: London Posts: 7 Thanks: 0 
yes de morgans is easy = not(p and q) == not p or not q not(p or q) == not p and not q 
May 12th, 2016, 07:15 PM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
$\displaystyle \neg(p\wedge q)\vee(\neg p\wedge q)$ $\displaystyle \equiv(\neg p\vee\neg q)\vee(\neg p\wedge q)$ $\displaystyle \equiv(\neg p\vee\neg q\vee\neg p)\wedge(\neg p\vee\neg q\vee q)$ $\displaystyle \equiv(\neg q\vee\neg p)\wedge(\neg p\vee1)$ $\displaystyle \equiv(\neg q\vee\neg p)\wedge1$ $\displaystyle \equiv(\neg q\vee\neg p)$ $\displaystyle \equiv(\neg p\vee\neg q)$ 
May 13th, 2016, 02:39 AM  #5 
Newbie Joined: May 2016 From: London Posts: 7 Thanks: 0 
Thanks Man! That has really helped me to understand. Do you know about this one. Given two statements, p and q, prove that (¬p ∧ q) → ¬ (q →p) is tautology. [Hint: p→q == (¬p ∨ q)] and verify your answer using a truth table. 
May 13th, 2016, 04:20 AM  #6 
Newbie Joined: May 2016 From: London Posts: 7 Thanks: 0 
(¬P∧ Q) → ¬ (Q → P) ¬ (¬P∧ Q) V ¬ (Q → P) (Implication Law) (¬¬ P V ¬Q) V ¬ (Q → P) (De Morgan’s Law) (P V ¬Q) V ¬ (Q → P) (Double Negation Law) (P V ¬Q) V ¬ (¬ Q V P) (Implication Law) (P V ¬Q) V (¬¬ Q ∧ ¬P) (De Morgan’s Law) (P V ¬Q) V (Q ∧ ¬P) = T (Double Negation Law) ¬ (¬P ∧ Q) V (Q ∧ ¬P) (Reversing De Morgan’s Law) ¬ (¬P ∧ Q) V (¬P ∧ Q) = T (Commutative Law) This Expression Reduces To True! P Q (¬P∧ Q) ¬ (Q → P) (¬P ∧ Q) → ¬ (Q →P) ¬ (¬P ∧ Q) V (¬P ∧ Q) F F F F T T F T T T T T T F F F T T T T F F T T This is what i have, but i am being told its wrong, help would be appreciated, thanks 
May 13th, 2016, 05:07 AM  #7  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  P Quote:
Next, you see that the expression on the right is a negation. Then there's a pretty good chance you'll need to assume the opposite at first, then prove that it's wrong by reductio. Now, (Q → P) is another conditional, so we can use the same strategy. Q is true according to (¬P∧ Q). If Q is true, we can derive P. But, according to (¬P∧ Q), we know that ~P is true as well. That contradicts what we've derived. Thus, (Q → P) is false. Then we go back and establish the validity of the whole thing, without dependencies. So here's the whole proof (left column shows the dependencies of each statement, and we've proved our tautology when we eliminate all the dependencies): 1 1. $\displaystyle (\neg P \wedge Q)$ (Assumption) 2 2. $\displaystyle (Q \rightarrow P)$ (Assumption) 1 3. $\displaystyle Q$ (1  Conjunction elimination) 1, 2 4. $\displaystyle P$ (2, 3  Conditional elimination) 1 5. $\displaystyle \neg P$ (1  Conjunction elimination) 1, 2 6. $\displaystyle (P \wedge \neg P)$ (4, 5  Conjunction introduction) 1 7. $\displaystyle \neg (Q \rightarrow P)$ (2, 6  Negation introduction) 8. $\displaystyle ((\neg P \wedge Q) \rightarrow \neg (Q \rightarrow P))$ (1, 7  Conditional introduction)  

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