# Prove De Morgan's LAw

by franz32
Tags: morgan, prove
 P: 133 I hope someone can help me prove one of De Morgan's Law: (A intersection B)' = A' U B'
HW Helper
P: 2,586
 Quote by franz32 I hope someone can help me prove one of De Morgan's Law: (A intersection B)' = A' U B'
I can suggest a way to prove it, although I've never formally done any of this stuff so I don't know if this approach is "acceptable." Anyways, it's a proof by contradiction. Assume the equation were false. That would mean that there exists and element, x, such that it is not in (A' U B') but is in (A intersection B)'. Now, if it is not in (A' U B'), then it is not in A' and it is not in B' (if it were in either of those, it would be in A' U B'). So, if it is not in A', it is in A, and if it is not in B', it is in B. Therefore, this element, x, is in A and it is in B, so it is in (A intersection B). Therefore, it is not in (A intersection B)'. This contradicts, the assumption, therefore the equation must be right.
 P: 618 franz32, Draw truth tables for both expressions. The expressions are equivalent if and only if their truth tables are the same.
 Emeritus Sci Advisor PF Gold P: 5,532 Prove De Morgan's LAw S=A U B means, "Every element of S is either an element of A or an element of B (or of both)", and T=A ^ B means, "Every element of S is an element of both A and B". We can write this using set builder notation: A U B={x|xεA or xεB} Take the complement: (A U B)'={x|xεA or εB}' Taking the complement of the statement on the RHS is the same as logically negating the "or" statement inside. Thus, (A U B)'={x|~(xεA or xεB)} Now apply DeMorgan's law for logical connectives to fit this to the definition of A'^B'. You can justify the use of DeMorgan's law for logic by using truth tables.
 Sci Advisor HW Helper P: 2,586 Proposition: $(A \cap B)' \subseteq (A' \cup B')$ Proof: (by contradiction)Assume: $(A \cap B)' \not\subseteq (A' \cup B')$ $\therefore \exists x \mbox{ such that } x \in (A \cap B)',\ x \notin (A' \cup B')$ $[x \notin (A' \cup B')] \Longrightarrow (x \notin A' \wedge x \notin B')$$\Longrightarrow (x \in A \wedge x \in B)$ $\Longrightarrow [x \in (A \cap B)]$ $\Longrightarrow [x \notin (A \cap B)']$ $\oplus \mbox{ (contradiction)}$$\therefore (A \cap B)' \subseteq (A' \cup B')\ \dots \ (1)$Proposition: $(A' \cup B') \subseteq (A \cap B)'$ Proof: (by contradiction)Assume: $(A' \cup B') \not\subseteq (A \cap B)'$ $\therefore \exists x \mbox{ such that } x \notin (A \cap B)',\ x \in (A' \cup B')$ $[x \notin (A \cap B)'] \Longrightarrow [x \in (A \cap B)]$$\Longrightarrow (x \in A \wedge x \in B)$ $\Longrightarrow [x \notin A' \wedge x \notin B')]$ $\Longrightarrow [x \notin (A' \cup B')]$ $\oplus \mbox{ (contradiction)}$$\therefore (A' \cup B') \subseteq (A \cap B)'\ \dots \ (2)$By (1) and (2), $(A' \cup B') = (A \cap B)'$
 P: 133 Oh.. I see Well, thanks for all your helps. I got it.
 Emeritus Sci Advisor PF Gold P: 11,155 Draw a Venn Diagram. It becomes painfully obvious.
Emeritus
PF Gold
P: 5,532
 Quote by Gokul43201 Draw a Venn Diagram. It becomes painfully obvious.
Ha! The last time I responded to a request for help in proving a set-theroetic statement, I said the same thing. But the poster said it wasn't allowed, so I did it "the hard way" this time.

It's true though, franz32, you can show that both statements have the same Venn diagram, and are therefore equivalent.
P: 948
 Quote by franz32 I hope someone can help me prove one of De Morgan's Law: (A intersection B)' = A' U B'
I haven't really looked at the LATEX stuff, so I'll have to do it with words.
Let x be in (A intersection B)'.
Then x is NOT in A intersection B
<=> x not in A AND x not in B (use logical connectors, negations, etc)
<=> x in A' OR x in B'
<=> x in A' U B'

That's one of them, hope it makes sense... you do the other one. (it's exactly the same)
 P: 95 doing a truth table would be difficult, but a venn diagram would be the least painful way. i know you have the answer but still for further problems like this, venn is the way to go.
Emeritus
PF Gold
P: 5,532
 Quote by 1+1=1 doing a truth table would be difficult,
It's no more difficult than the Venn diagram. Once you reduce DeMorgan's law for sets to DeMorgan's law for logic, all you have to do is show that ~(p and q) has the same truth table as (~p or ~q).

~(p and q)
Since (p and q) is true only when (p,q)=(T,T), the negation is false only when (p,q)=(T,T).

(~p or ~q)
Since (p or q) is only false when (p,q)=(F,F), the statement (~p or ~q) is false only when (p,q)=(T,T).

Since the two statements have the exact same truth table, they are equivalent.
 P: 1 hello can some one Plz explain to me my task: Prove De Morgan Law with the appropriate explanation were p and q are sentence meaning (proposition) ~(p ^ ( it s OR i dnt knw how to type on comuter) ---(if) (~p) ^ (~q) and which font should i use so that i can type the symbols and if some one will prove it i will be glad
Emeritus