Chandasouk
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Let A, B, and C be sets. Show that
a) (A-B) - C [itex]\subseteq[/itex] A - C
b) (B-A) [itex]\cup[/itex] (C-A) = (B [itex]\cup[/itex] C) - A
I am using variable x to represent an element.
Part A)
I rewrote (A-B) - C as (x[itex]\in[/itex]A ^ x[itex]\notin[/itex]B) - C
I think this could be rewritten as
(x[itex]\in[/itex]A ^ x[itex]\notin[/itex]B) ^ x[itex]\notin[/itex] C
A-C can be rewritten as (x [itex]\in[/itex] A ^ x [itex]\notin[/itex] C)
The original statement can be rewritten as
x[itex]\in[/itex]A[itex]\cap[/itex]~B[itex]\cap[/itex]~C [itex]\subseteq[/itex] x[itex]\in[/itex]A[itex]\cap[/itex]~C
where ~ represents negation.
However, for the LHS to be a subset of the RHS, all elements of the LHS should be an element of RHS but since the LHS has ~B, I don't think that it is a subset?
I have no idea how to show part B so any help would be great.
a) (A-B) - C [itex]\subseteq[/itex] A - C
b) (B-A) [itex]\cup[/itex] (C-A) = (B [itex]\cup[/itex] C) - A
I am using variable x to represent an element.
Part A)
I rewrote (A-B) - C as (x[itex]\in[/itex]A ^ x[itex]\notin[/itex]B) - C
I think this could be rewritten as
(x[itex]\in[/itex]A ^ x[itex]\notin[/itex]B) ^ x[itex]\notin[/itex] C
A-C can be rewritten as (x [itex]\in[/itex] A ^ x [itex]\notin[/itex] C)
The original statement can be rewritten as
x[itex]\in[/itex]A[itex]\cap[/itex]~B[itex]\cap[/itex]~C [itex]\subseteq[/itex] x[itex]\in[/itex]A[itex]\cap[/itex]~C
where ~ represents negation.
However, for the LHS to be a subset of the RHS, all elements of the LHS should be an element of RHS but since the LHS has ~B, I don't think that it is a subset?
I have no idea how to show part B so any help would be great.