Prove that if A intersect B = A intersect C then

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SUMMARY

The discussion centers on proving the equivalence A ∩ B = A ∩ C if and only if A ∩ ¬B = A ∩ ¬C. Participants are struggling with the proof and have attempted various approaches, including manipulating set operations and using element membership. The key steps involve demonstrating that if an element x belongs to A and either B or C, it must also belong to the complements of B and C under the intersection with A. The proof requires a clear understanding of set theory operations and logical implications.

PREREQUISITES
  • Understanding of set theory, specifically intersection and complement operations.
  • Familiarity with logical equivalences and implications in mathematical proofs.
  • Knowledge of element membership notation (e.g., x ∈ A).
  • Basic skills in manipulating algebraic expressions involving sets.
NEXT STEPS
  • Study the properties of set operations, focusing on intersection and complement.
  • Learn about logical equivalences and how to apply them in proofs.
  • Practice proving set identities using element membership arguments.
  • Explore examples of similar proofs in set theory to reinforce understanding.
USEFUL FOR

Students of mathematics, particularly those studying set theory and logic, as well as educators looking for examples of proofs involving set operations.

Andrax
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Homework Statement



Prove that if A [itex]\cap[/itex] B = A [itex]\cap[/itex] C [itex]\Leftrightarrow[/itex] A [itex]\cap[/itex] [itex]\neg[/itex]B = A [itex]\cap[/itex] [itex]\neg[/itex]C

Homework Equations


The Attempt at a Solution


I have done a lot of work on this exercise nothing seems to work..
 
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Andrax said:
I have done a lot of work on this exercise nothing seems to work..

Can you show us the work you did?
 
All i pretty much did so far was turning the A [itex]\cap[/itex] LB = A [itex]\cap[/itex] C [itex]\Leftrightarrow[/itex] A - B = A - C
i've tried pretty much everything nothing seems to work
example : let x E a ---> x E A [itex]\cap[/itex] B ---> x E A [itex]\cap[/itex] C
yeah I'm pretty much stuck here
also i tried this one A [itex]\cap[/itex] B = A [itex]\cap[/itex] C [itex]\longrightarrow[/itex] x E A and x E B and x E C [itex]\Longrightarrow[/itex] x E A and x E l B and x E l C And I am stuck here.. i need to prove that x E A and x E/ B = x E A and x E/ C...
 

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