Suggestion: Just write your proof in a single post rather than asking people to click on a different URL for every single line in your proof.
Can someone check if my working is correct for the following proof?
Prove for all sets and .
Let be the proposition function '' and be the proposition function ''
Let the domain of discourse of both these proposition functions be , the universal set.
Now to prove the statement to be true we must show the following 2 universally quantified statements to be true:
1.
2.
To show case 1 to be true, first we know that if is false then the case is vacuously true, so we will ignore this trivial case.
We will assume to be true and if we can show that is also true then case 1 is true.
Now and
Since is true then either
a. is true and is false
b. is false and is true
[Note: these 2 propositions can not be both true since they are mutually exclusive]
If we use case a. Since is true is true is true.
Now using case b. Since is true is true is true is true.
Thus we have shown case 1. to be true.
To show case 2 to be true, first we know that if is false then the case is vacuously true, so again we will ignore this trivial case.
We will assume Q(x) to be true and Q(x) is true if either of the following 3 conditions are satisfied.
i. is true and is false
ii. is false and is true
iii. is true and is true
Now using case i. Since is true is true is true.
Using case ii. Since is true is true is true is true.
Now using case iii. Since is true and is true is true is true.
Case 2 is now proven to be true.
We have completed our proof to show that the original statement is true.
Thanks!
I don't see Latex in your post. In other poster's posts I see Latex, but for your post, my browser shows virtually only English words plus URLs.
But another poster does see your Latex, so I guess it's an anomaly between your post and my broswer.
Anyway, it's very simple.
Show X u (Y\X) = X u Y:
Suppose z in X u (Y\X).
So z in X or z in Y\X.
If z in X, then z in X u Y.
If z in Y\X, then z in X u Y.
Suppose z in X u Y.
So z in X or z in Y.
Suppose z in X.
So z in X u (Y\X).
Suppose z in Y.
If z in X, then see above.
If z not in X, then Z in Y\X, so z in X u Y\X.
So z in X u (Y\X) iff X u Y.
So, since z is arbitary, we have X u (Y\X) = X u Y.
/
Even more brief:
z in X u Y
iff
z in X or z in Y
iff
z in X or ((z in Y but z not in X) or (z in Y and z in X))
iff
z in X or (Z in Y but z not in X)
iff
z in X or z in Y\X
iff
z in X u Y\X
/
Even more brief:
z in X u Y
iff
z in X or z in Y
iff
z in X or (z in Y but z not in X)
iff
z in X u Y\X
/
Even more brief:
z in X u Y
iff
z in X or (z in Y but z not in X)
iff
z in X u Y\X
/
Even more brief:
Follows by obvious sentential logic and universal generalization.