# Thread: Two proofs I've done...

1. ## Two proofs I've done...

Hi guys I did two proofs for a logic class that I'm in. Could someone please check these to make sure I'm on the right track?

For all sets A and B, (A\B)U(AnB)=A.

Let x E A
if x E B, then x E (AnB)
if x not E B, then x E (A\B)
therefore x E (A\B)U(AnB)
therefore (A\B)U(AnB) = A

Let S be a subset of a universal set U. The characteristic function fs is the function from U to the set {0,1} defined by

fs(x)={1, if x E S
{0, if x not E S
Let A and B be sets. Show that for all x, fAnB(x) = fA(x) * fB(x)

Let x E U
Case 1: If x E A ^ x E B, then fsAnB(x) = 1 = 1*1 = fsA(x) * fsB(x)
Case 2: If x E A but x not E B, then fsAnB(x) = 0 = 1*0 = fsA(x) * fsB(x)
Case 3: If x not E A but x E B, then fsAnB(x) = 0 = 0*1 = fsA(x) * fsB(x)
Case 4: If x not E A ^ x not E B, then fsAnB(x) = 0 = 0*0 = fsA(x) * fsB(x)

2. Originally Posted by Benderbrau
Hi guys I did two proofs for a logic class that I'm in. Could someone please check these to make sure I'm on the right track?

For all sets A and B, (A\B)U(AnB)=A.

Let x E A
if x E B, then x E (AnB)
if x not E B, then x E (A\B)
therefore x E (A\B)U(AnB)
therefore (A\B)U(AnB) = A
No, you only proved that that A is contained in the union. You have to show the reverse direction.
Originally Posted by Benderbrau

Let S be a subset of a universal set U. The characteristic function fs is the function from U to the set {0,1} defined by

fs(x)={1, if x E S
{0, if x not E S
Let A and B be sets. Show that for all x, fAnB(x) = fA(x) * fB(x)

Let x E U
Case 1: If x E A ^ x E B, then fsAnB(x) = 1 = 1*1 = fsA(x) * fsB(x)
Case 2: If x E A but x not E B, then fsAnB(x) = 0 = 1*0 = fsA(x) * fsB(x)
Case 3: If x not E A but x E B, then fsAnB(x) = 0 = 0*1 = fsA(x) * fsB(x)
Case 4: If x not E A ^ x not E B, then fsAnB(x) = 0 = 0*0 = fsA(x) * fsB(x)
Yes this is valid since you enumerated all cases.

3. So in order to complete the first proof, could I do something like this?

Let x E (A\B)U(AnB)
if x E (A\B), then x not E B
if x E (AnB), then x E B
therefore x E A

4. Originally Posted by Benderbrau
So in order to complete the first proof, could I do something like this?

Let x E (A\B)U(AnB)
if x E (A\B), then x not E B
if x E (AnB), then x E B
therefore x E A
The steps are correct, but the two then's are wrong. You should conclude $x\in A$ in both cases.