I don't understand the transformation of this logic statement

bushra1175

Junior Member
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Jun 14, 2020
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59
Here are the laws in the textbook:
1613942693385.png

The distributive law was applied to the upper statement, resulting in the lower one:
1613942801877.png

I have stared at this for a long time and the transformation looks completely unrelated to the distributive law illustrated in the texbook. This snipet is part of a simplification of the below logic statement. I understand everything up until the highlighted part:

1613943080719.png
1613943169009.png
 
Replace the expressions I've highlighted with a new name, r. Does it look like distribution then?

1613957540357.png

You may have to use mirror vision -- distribution works from either side.
 
Replace the expressions I've highlighted with a new name, r. Does it look like distribution then?

View attachment 25251

You may have to use mirror vision -- distribution works from either side.

I can see that would make it

= (¬p ∧ q) v r
= [¬p v r] ∧ [ q v r]


and since (¬p ∧ q) v r matches with 1613963093672.png , then [¬p v r] ∧ [ q v r] should match with 1613963135736.png, but it doesn't

Instead, the result [¬p v r] ∧ [ q v r] matches with the other distributive law 1613963233023.png

Why is this the case?
 
I can see that would make it

= (¬p ∧ q) v r
= [¬p v r] ∧ [ q v r]


and since (¬p ∧ q) v r matches with View attachment 25254 , then [¬p v r] ∧ [ q v r] should match with View attachment 25255, but it doesn't

Instead, the result [¬p v r] ∧ [ q v r] matches with the other distributive law View attachment 25256

Why is this the case?

Because it's law 15, commuted! As I said, it works from either side -- but you have to turn everything around. What's inside the parentheses is far more important that the order of the pieces, and you're seeing it wrong.

Law 15, written reversed, says \((q\wedge r)\vee p = (q\vee p)\wedge(r\vee p)\): 1613966406258.png

This is just like the fact that the distributive property with numbers can be written as \(a(b+c) = ab + ac\) or as \((b+c)a = ba + ca\).
 
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