There is a subset of set P, the set of people, which is always a subset of set F, the set of fooled people.

There are times when the set P and set F are equivalent.

**EDIT:**It's Friday evening, so this is still today's strip. I updated the chart. It didn't seem right to me last night, and I tweaked it. I just didn't tweak it correctly. Oopsie. Lincoln must've been better at this than I am.

## 3 comments:

A ⊂ A ∩ B ==> a ≡ B...

perhaps you meant A ⊃ A ∩ B

A = {people }

B = {x| fooled(x)}

Could be.

I threw it the set stuff at the last minute. The graph was my original idea. I thought it needed something else to it.

Now if you'll excuse me, I'll have to look up what you wrote just to make sure I'm following. Teach Algebra for 8 years (after JHS math for 2), and some of it starts to slip away from you.

I am unable to understand your query but i like to

Please provide simpler solution for easy assistance just like Abacusprovide for maths calculation.

Waiting for your reply..

Post a Comment