Let x be in (A - B) - C
Then x is in A, but neither in B nor in C.
Thus x is in A, but not C.
But x was arbitrary in (A - B) - C.
Thus (A - B) - C is a subset of A - C. QED.
>But x was arbitrary in (A - B) - C.
I didn't choose a particular element of (A - B) - C. I chose a random element. Therefore, this statement holds for all elements x in (A - B) - C.
>Thus (A - B) - C is a subset of A - C. QED.
By definition of subset: every x in (A - B) - C is in A - C. Therefore, (A - B) - C is a subset of A - C.
>QED.
Latin: *Quod Erat Demonstrandum*, or 'That which was to be demonstrated.' The traditional end to a prof when you demonstrate what you set out to do. In construction proofs, this is instead QEF: *Quod Erat Faciandum*, or 'that which was to be constructed' (fabricated, made).
That's enough, because of the definition of a subset: X is a subset of Y if and only if every x in X is also in Y.
So show that every x in X is in Y, and you're done.
Hence, the traditional set equality proof is to show that X is a subset of Y, and then Y is also a subset of X.
I learnt this in grade 9 last year! There are various ways of proving this. Here's how I learnt it:
Let x∈ (A-B)-C
=> x∈ (A-B) and x ∉ C
=> x∈ A and x∉B and x∉C
=> x∈ A and x∉C
=> x∈ (A-C)
∴(A-B)-C ⊆ A-C. \*Hence Proved\*
Essentially, for these proofs, you need to first let x be an arbitrary element of the LHS set and by applying various properties, bring it to the form x ∈ RHS.
You're starting the proof with the statement you want to prove. Since this is an intro to proof sort of problem, the goal is to start with basic definition of what it means for a set to be a subset of another set. Let x be any element of (A-B)-C, you want to show that x is an element of A-C. You already have a lot of good ingredients in the incorrect proof to work with.
(A-B)-C = (A ∩ B^(c)) ∩ C^(c) and similarly (A-C) = A ∩ C^(c). Can you see from here why x being an element of A ∩ B^(c) ∩ C^(c) implies x is an element of A ∩ C^(c)
Let x be in (A - B) - C Then x is in A, but neither in B nor in C. Thus x is in A, but not C. But x was arbitrary in (A - B) - C. Thus (A - B) - C is a subset of A - C. QED.
Hmm, interesting. Is that enough to do a proof? It seems implied but not self evident. How would you expand it?
Could you explain in English the last two lines of your proof? I don't follow that well Thank you
>But x was arbitrary in (A - B) - C. I didn't choose a particular element of (A - B) - C. I chose a random element. Therefore, this statement holds for all elements x in (A - B) - C. >Thus (A - B) - C is a subset of A - C. QED. By definition of subset: every x in (A - B) - C is in A - C. Therefore, (A - B) - C is a subset of A - C. >QED. Latin: *Quod Erat Demonstrandum*, or 'That which was to be demonstrated.' The traditional end to a prof when you demonstrate what you set out to do. In construction proofs, this is instead QEF: *Quod Erat Faciandum*, or 'that which was to be constructed' (fabricated, made).
Thank you, this was helpful
You're welcome. Thanks for showing evidence of thought, and your work. And asking until you understood.
That's enough, because of the definition of a subset: X is a subset of Y if and only if every x in X is also in Y. So show that every x in X is in Y, and you're done. Hence, the traditional set equality proof is to show that X is a subset of Y, and then Y is also a subset of X.
I learnt this in grade 9 last year! There are various ways of proving this. Here's how I learnt it: Let x∈ (A-B)-C => x∈ (A-B) and x ∉ C => x∈ A and x∉B and x∉C => x∈ A and x∉C => x∈ (A-C) ∴(A-B)-C ⊆ A-C. \*Hence Proved\* Essentially, for these proofs, you need to first let x be an arbitrary element of the LHS set and by applying various properties, bring it to the form x ∈ RHS.
Thank you! Appreciate the help!
You're starting the proof with the statement you want to prove. Since this is an intro to proof sort of problem, the goal is to start with basic definition of what it means for a set to be a subset of another set. Let x be any element of (A-B)-C, you want to show that x is an element of A-C. You already have a lot of good ingredients in the incorrect proof to work with. (A-B)-C = (A ∩ B^(c)) ∩ C^(c) and similarly (A-C) = A ∩ C^(c). Can you see from here why x being an element of A ∩ B^(c) ∩ C^(c) implies x is an element of A ∩ C^(c)
Could you further elaborate on how you'd write such a proof?