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..
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