#Set proof; Don’t really know where to go and hit a brick wall
bold pantherBOT
coral girder
it might help to consider the contrapositive, instead of showing that x not in D -> x in B, it would be easier to show that x not in B -> x in D.
tawdry wren
it might help to consider the contrapositive, instead of showing that x not in D...
Ok,I’ll take a look. Last time I did a proof by contrapositive,I had the mishap of messing up the proof; So may I get your help with properly negating it?
My understanding:
P —> Q
CONTRAPOSITIVE:
-Q —> -P
coral girder
yes
that's correct
they're equivalent statements, so if showing P -> Q is too hard, it's equally powerful to show ~Q -> ~P
and in your example, the contrapositive would be x not in B -> x in D, agreed?
tawdry wren
and in your example, the contrapositive would be x not in B -> x in D, agreed?
Yes
And,with the new problem,How can we use that leftmost piece of information in the picture? That part just stumps me
I don’t see how to follow through with it 🛁
coral girder
okay, try assuming x isn't in B since we're trying to show the contrapositive
we know from the problem that x is in A, and we're assuming that x isn't in B
what might this tell you about how x relates to A \ B?
tawdry wren
what might this tell you about how x relates to A \ B?
Does this mean we double negate the “x -€ B” in this?
coral girder
no, do you know what the notation A \ B means?
tawdry wren
no, do you know what the notation A \ B means?
Using the connectives,you would get “x € A AND x -€ B”
The negative is there for negation
coral girder
Using the connectives,you would get “x € A AND x -€ B”
yes, so these two things are true. and A \ B contains exactly the elements in A that aren't in B
do you see why this would mean that x is an element of A \ B?
tawdry wren
Not really…
coral girder
by definition, the set A \ B = {x in A | x not in B}
it's the set difference, in other words, it's the elements of A that aren't included in B
tawdry wren
Yes
coral girder
we know from the problem that x is in A, and from assuming the contrapositive, x isn't in B
coral girder
so you agree with me that x is an element of A \ B?
tawdry wren
Yes
coral girder
if A \ B is a subset of C intersect D, every element of A \ B is contained in C intersect D
and we know x is in A \ B
what would this tell you about how x relates to C intersect D?
tawdry wren
That means that x is an element of C intersect D
Writing out the definition of this
That means that x is an elect of C and x is an element of D
coral girder
yes, so you've shown that x would be in D
so big picture what we did, we wanted to show that x not in D implies x in B.
by contrapositive, we can show that x not in B implies x in D
from x in A and x not in B, we know x in A \ B
since A \ B is a subset of C intersect D, we know x in C intersect D
thus x in C and x in D, so we showed that x in D which is the contrapositive statement we wanted to prove
tawdry wren
So we’ve proved it?
It makes so much sense now…I can’t believe I didn’t see it before
Thank you so much for your help,I haven’t been sure of how to approach this problem until now
severe crescent