Question

Math proof. Can some here actually help! Thanks.

Answer 1

typing it up

Answer 2

\[\cup _ i \in I (A_i \cap B_i) \subseteq (\cup_i \in I A_i)\cap(\cup_i \in I B_i)\]

Answer 3

the in I parts should be sub scripted with the respective unions

Answer 4

$$ \large{
\cup _ {i \in I }(A_i \cap B_i) \subseteq (\cup_{i \in I}~ A_i)\cap(\cup_{i \in I}~ B_i)
}$$

Answer 5

thanks

Answer 6

Also this works
$$ \large{
\bigcup _ {i \in I }~(A_i \cap B_i) \subseteq (\bigcup_{i \in I}~ A_i)\cap(\bigcup_{i \in I}~ B_i)
}$$

Answer 7

Okay thanks, so how can I prove this?

Answer 8

show that for an arbitrary x, if x is an element of the left side, it is also an element of the right side

Answer 9

Yes, but I need to use logical forms to show that left side is a subset of right side

Answer 10

it needs to be a rigorous proof

Answer 11

try expanding the left side, use

Answer 12

$$ \large \rm {

\bigcup _ {i \in I }~(A_i \cap B_i) = (A_1 \cap B_1) \cup (A_2 \cap B_2) \cup ... (A_n \cap B_n)
}$$

Answer 13

even though \( I \) does not have to be a countable set, but this is a good start to get a feel where to go from here

Answer 14

$$ \rm {

\bigcup _ {i \in I }~(A_i \cap B_i) = (A_1 \cap B_1) \cup (A_2 \cap B_2) \cup ... (A_n \cap B_n)
}
$$

Answer 15

I did that but I don't know what to do after that

Answer 16

I got both sides to expanded form

Answer 17

we can apply distributive laws to the expanded form

Answer 18

my teacher wants the logical equivalent to these things though

Answer 19

and using laws with the logical forms

Answer 20

\[\cup_{i \in I}(A_i \cap B_i) \leftrightarrow (x \in A_1 \land x \in B_1) \lor ....( x \in A_i \land x \in b_i)\]

Answer 21

ok we can do that

Answer 22

Okay please show me the path lol

Answer 23

we are dealing with a lot of statements here , but

Answer 24

$$ x \in \cup_{i \in I}(A_i \cap B_i) \iff \\
(x \in A_1 \land x \in B_1) \lor ( x \in A_2 \land x \in B_2) \lor ...( x \in A_n \land x \in B_n)

$$

Answer 25

lets look at the first two statements
( p & q ) V ( r & s ) , how can we distribute that

Answer 26

distributive law

Answer 27

( p & q ) V ( r & s )
= (( p & q ) V r ) & ( (p &q ) v s )

Answer 28

is p=a1 and b1

Answer 29

right

Answer 30

( p & q ) V ( r & s )
= (( p & q ) V r ) & ( (p &q ) v s )
= (r v ( p & q ) ) & ( s v (p &q ) )
= (( r v p ) & ( r v q )) & (( s v p ) & (s v q))

you can remove the outermost parentheses

Answer 31

okay just to clarify: is p =a1 and q=b1 or is p=a1 and b1
and the what is q ?

Answer 32

p = a1 , q = b1
r = a2, s = b2

Answer 33

I don't think distributive law works with four terms only 3

Answer 34

P^(QvR) = (P^Q)v(P^R)

Answer 35

at least thats what I have in my notes

Answer 36

i treated p &q as one term first

Answer 37

so p and q = x or something?

Answer 38

do you see how i did the first step
( p & q ) V ( r & s )
= (( p & q ) V r ) & ( (p &q ) v s )

Answer 39

now, after looking at it again yes. thanks.
So why can you take off the parenthesis on the outer part

Answer 40

for the last step?

Answer 41

yes

Answer 42

by the associativity rule. you can drop parentheses if you have all &'s

Answer 43

oh I didn't know that.

Answer 44

carefully though

Answer 45

( p & q ) V ( r & s )
= (( p & q ) V r ) & ( (p &q ) v s )
= (r v ( p & q ) ) & ( s v (p &q ) )
= (( r v p ) & ( r v q )) & (( s v p ) & (s v q))
= ( r v p ) & ( r v q ) & ( s v p ) & (s v q)
= ( p v r ) & ( q v r ) & (p v s ) & ( q v s )

Answer 46

okay I see. is this the last step? plugging in for all values looks kinda confusing

Answer 47

i used commutativity ,
r v p = p v r

Answer 48

yeah but I don't see what the end result of all of this should be. Do we need to do the same for the right hand side to show that both sides are equal now ?

Answer 49

i have to write this on paper, let me see

Answer 50

okay sure

Answer 51

$$
x \in \cup_{i \in I}(A_i \cap B_i) \\ \iff \\
(x \in A_1 \land x \in B_1) \lor ( x \in A_2 \land x \in B_2) \lor ...( x \in A_n \land x \in B_n)
\\ \iff \\
\\ (x \in A_1 \lor x \in A_2 \lor .. x \in A_n) \land ( x \in B_1 \lor x \in B_2... \lor~ x \in B_n)


$$

Answer 52

Is that last step just
= ( r v p ) & ( r v q ) & ( s v p ) & (s v q)
= ( p v r ) & ( q v r ) & (p v s ) & ( q v s )

Answer 53

that was commutativity. i wanted to show that

Answer 54

Claim: ( p & q ) V ( r & s ) = ( p v r ) & ( q v r ) & (p v s ) & ( q v s )

Proof:
( p & q ) V ( r & s )
= (( p & q ) V r ) & ( (p &q ) v s )
= (r v ( p & q ) ) & ( s v (p &q ) )
= (( r v p ) & ( r v q )) & (( s v p ) & (s v q))
= ( r v p ) & ( r v q ) & ( s v p ) & (s v q)
= ( p v r ) & ( q v r ) & (p v s ) & ( q v s )

Answer 55

Okay great I see now! one last thing: is this a rigor proof since we don't check all of the the index instead just the first few terms?

Answer 56

i think we could simplify a bit more

Answer 57

how so

Answer 58

I am trying to use wolram to show that the two expressions are equal

Answer 59

wolfram

Answer 60

oh thats okay I can see that they would be, but I was just wondering if this is a rigorous proof

Answer 61

not completely rigorous

Answer 62

how can I make it completely rigorous

Answer 63

Thanks a lot for your help perl I appreciate it