solidworks in wine... not yet
Check out the wine app db:
not quite there yet: extruding crashes it
http://appdb.winehq.org/objectManager.php?bShowAll=true&bIsQueue=false&bIsRejected=false&sClass=version&sTitle=&sReturnTo=&iId=8983
For instance, the following seems obvious but I don't know how to state the proof formally (and directly):
Show A \cap (B-A) = \{\}
Here is a try:
For any x \in U \ if \ x \in A then x \notin (B-A)
therefore A \cap (B-A) = \{\}
there is something missing...
Here is a go at a direct proof:
A\cap B = \{\} \Leftrightarrow \forall x \in U \ \ x \notin A\cap B
\Leftrightarrow x \in (A\cap B)^c
\Leftrightarrow x \in A^c\cup B^c
Now suppose x \in A then x \in B^c and
\Leftrightarrow A \subseteq B^c
is this correct? is this what you meant by...
Hi. I am new to formal proofs. Is the following a legitimate method for proving this assertion by contradiction.
Show A\cap B = \{\} \Leftrightarrow A \subseteq B^c
If A\cap B \neq \{\} then \exists \ x \ ST \ x \in A \ AND \ x \in B
Since x \in A \ \ \ A \subseteq B^c...
Hi,
Why is it that if A is m×n-matrix and B is n×m matrices such that m<n, then AB is m×m and BA is n×n matrix. Then the following is true:
pAB(t) = t^(m-n)*pBA(t)
where pAB(t) and pBA(t) are characteristic polynomials of AB and BA
thanks