lec0125 - A Note of variable substitution In each of the...

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A Note of variable substitution In each of the formulas I have gone over in chapters 2 and 3, each variable is exactly that, a variable. A variable is something that you can substitute in any value of the similar type for...that is what makes formulas with variables so powerful – they are applicable to a large number of cases. Here is an example of what I mean: The Distributive Law is: p (q r) (p q) (p r) Imagine simplifying the following expression: ((s t) v) ( ¬ v (s t)) ((s t) v) ((s t) ¬ v), Commutative (s t) (v ¬ v), Distributive, with p= s t, q= v, r= ¬ v (s t) F, Inverse Law (s t) , Identity Law One other thing to mention: notice how I used different letters in the problem than are listed in the rules...The reason I did that was for clarity’s sake; to show you all what was substituted for each variable. BUT, in the problems that YOU do, you’ll find that ps, qs and rs are used, just like in the formula. When you make substitutions, you’ll have to differentiate between the p&qs from the formula, and the p&qs in your specific problem!!! Be careful when you are doing this.
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Last lecture we ended with the problem of showing the following: ¬ ( ¬ ((A B) C) ¬ B) = (B C) One way is to use a membership table: A B C B C ¬ ((A B) C) ¬ B ¬ ( ¬ ((A B) C) ¬ B) 0 0 0 0 1 1 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 0 1 1 1 1 0 0 1 Next, we could use Set Laws to show the two expressions to be equivalent: ¬ ( ¬ ((A B) C) ¬ B) = ¬¬ (( A B) C) ¬¬ B (De Morgan’s) = (( A B) C) B (Double Negation) = (( A B) B) C (Associate & Commutative) = B C (Absorption) For this particular problem, we could show that each set is a subset of the other. However, for this particular problem, that is rather tedious. I will give other examples of this nature to
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