Question

Verify that the following are tautologies by citing the appropriate previous result. Do not make truth tables.

Answer 1

Hmmm, are you going to draw your work like you always do?

Answer 2

just 1.4.15 I've done 1.4.16 so anyway if I draw the truth tables I have seen the column of T's but I can't do that on 1.4.15

Answer 3

and the cite appropritate previous results is driving me nuts.

Answer 4

I feel like the problem wants you to cite the theorems that the textbook establishes

Answer 5

I've noticed that a. is a double negation b and c are demorgans d and e is divisibility

Answer 6

the last one is contrapositive.

Answer 7

you can't cite problem 1.4.12 for f?

Answer 8

o_O what is F and I wasn't assigned 1.4.12

Answer 9

wait a second...
http://en.wikipedia.org/wiki/Material_implication_(rule_of_inference)

Answer 10

I think you are supposed to use various properties of Boolean Algebra, no?

Answer 11

what is that? D: like communitative associative law?

Answer 12

Yes.

Answer 13

maybe from this?

Answer 14

I think F is a combination of material implication/rule of inference, and de morgan's law after negation

Answer 15

ok so how do I verify a tautology if I have to use those laws?

Answer 16

Yes, you can use the double negative property to show that \[
\neg(\neg P)\iff P
\]Is equivalent to \[
P \iff P
\]

Answer 17

Crap what happened to my latex?

Answer 18

so I have to use those laws and show equivalency?

Answer 19

Why can't I see previews either?

Answer 20

I have texmaker

Answer 21

~(p->q) <-> ~(~p v q) Material implication
~(~p v q) <-> (~~p)^(~q) De Morgan's Law
(~~p)^(~q) <-> p^(~q) Double Negation

Answer 22

so using demorgans and those laws....

Answer 23

~(~p) <-> p double negation

Answer 24

then apply demorgans to that/

Answer 25

so...
~[~(~P)] <-> ~P

Answer 26

:/

Answer 27

ohhhhh.
~(~P) <-> ~(~P)
P <-> P

Answer 28

I did f for you, but I'm not sure if your textbook has the material implication/rule of inference. It should though...

Answer 29

I'm not sure if I'm doing a right... I was just substituting...
~(~P) <-> P

~(~P) <-> ~(~P)

P <-> P

Answer 30

@UsukiDoll , are you citing the laws you used?

Answer 31

EEP!
~(~P) <-> P is double negation and then

Answer 32

I was subsituting stuff afterwards.......
no good... gotta use demorgans on a

Answer 33

huh?

Answer 34

@______________@

Answer 35

well wio put that ~(~P) <-> P
is equilivlent to P <-> P

Answer 36

yeah

Answer 37

the latex on here wasn't working, but I used texmaker to see what it was

Answer 38

maybe ~(~P) <-> P
is equilvalent to ~(~P) <-> ~(~P)?

Answer 39

P <-> P
and then use a textbook theorem to show it is true

Answer 40

so I can't use ~(~P) <-> ~(~P) ?

Answer 41

Well, I'm not sure why you would complicate things...
~(~P) <-> P
is equivalent to P <-> P by double negation and thus is T

Answer 42

not sure which law it is that states that P<->P is a tautology but I'm sure there is a name for it.

Answer 43

did you just subsitute the ~(~P) with the P?
for ~(~P) <-> P?

Answer 44

yes, and my justification for that is the double negative

Answer 45

the main problem I see with using the <-> both to express the expression and to show equivalence is that it winds up getting kinda confusing

Answer 46

Since \(x=x\) is a reflexive property of equality, you'd cite the reflexive property of the bi-conditional operator.

Answer 47

If you want to be that rigorous.

Answer 48

<-> is confusing... there are two meanings.. one is biconditional and the other means equilvalent... the one in the book at least for this problem has a biconditional arrow

Answer 49

yeah, my teacher usually uses three bars

Answer 50

P <-> P is A tautology since P has the truth table of T T F F
another P would be T T F F
P <-> P is T T T T

Answer 51

~P is F F T T
~(~P) is T T F F
another ~(~P) is T T F F
so ~(~P) <-> ~(~P)
is T T T T

Answer 52

yeah - my teacher would probably just let me stop at p<->p since he would consider the fact that that is true trivial

Answer 53

alright 2 down 4 more of those bad boys... luckily b and c are demorgans... and d and e are distributives

Answer 54

so... for b we have...
~( P V Q) <-> ~P ^ ~Q

Answer 55

maybe it's equilvalent to
ahhh screw spelling lol
~P ^ ~Q <-> ~P ^ ~Q

Answer 56

and ~( P V Q) <-> ~( P V Q)

Answer 57

-_- this would be a lot easier if I draw the table slightly.
P is T T F F
Q is T F T F
P V Q is T T T F
~(P V Q) is F F F T
another ~(P V Q) is F F F T
~( P V Q) <-> ~( P V Q) is T T T T

Answer 58

~P ^ ~Q <-> ~P ^ ~Q
P is T T F F
Q is T F T F
~P is F F T T
~Q is F T F T

~P ^ ~Q is F F F T
another ~P ^ ~Q is F F F T
~P ^ ~Q <-> ~P ^ ~Q is T T T T

Answer 59

c. ~(P ^ Q) <-> ~P V ~Q

so... ~P V ~Q <-> ~P V ~Q
~(P ^ Q) <->~(P ^ Q)

Answer 60

P is T T F F
Q is T F T F
~P is F F T T
~Q is F T F T
~P V ~Q is F T T T
another ~P V ~Q is F T T T
~P V ~Q <-> ~P V ~Q is T T T T

Answer 61

ugh too sleepy to look over your work; @wio are you there?

Answer 62

P is T T F F
Q is T F T F
~P is F F T T
~Q is F T F T
(P ^ Q) is T F F F
~(P ^ Q) is F T T T
another ~(P ^ Q) is F T T T
~(P ^ Q) <->~(P ^ Q)
T T T T

Answer 63

maybe that's how it's supposed to be done?

Answer 64

If it is... then I'll just do D and E on paper since that's 8 rows right there

Answer 65

Aren't you not supposed to use truth tables here?

Answer 66

yeah *facepalm*

Answer 67

arghhhhhhhhhhhhh ...... I know b and c are demorgans...

Answer 68

so a demorgan within a demorgan

Answer 69

just a single de morgan?

Answer 70

b is one demorgan and c is another demorgan

Answer 71

D: curse the instructions! I WISH I COULD USE A TRUTH TABLE!

Answer 72

~( P V Q) <-> ~P ^ ~Q

Answer 73

yeah, that's just de morgans law one time?

Answer 74

and then c. was the other demorgan c. ~(P ^ Q) <-> ~P V ~Q

Answer 75

@eliassaab

Answer 76

demorgan within a demorgan...................................

Answer 77

Well, de Morgan is used regardless.

Answer 78

but I can't just leave the questions blank...

Answer 79

welll for those one step problems just cite the rule, and for multi step problems just do a mini proof I guess?

Answer 80

b and c are just demorgans... I guess that's the answer?

Answer 81

maybe do something like this?
http://math.stackexchange.com/questions/358750/prove-the-following-is-a-tautology

Answer 82

Not sure what your teacher wants, but mine usually liked my two column proofs - anythign clear I guess

Answer 83

I wish it was a truth table... that would've been easy

Answer 84

hmmmmmm b and c are demorgans
d and e are distributives
g is a contra positive

Answer 85

~( P V Q) <-> ~P ^ ~Q demorgan

then another demorgan usage
demorgan tautology end result the end

Answer 86

wait the other demorgan is in c...
maybe...
~( P V Q) <-> ~P ^ ~Q Demorgan's law
~(P ^ Q) <-> ~P V ~Q Demorgan's Law

Answer 87

and maybe for c
~(P ^ Q) <-> ~P V ~Q Demorgan's Law
~( P V Q) <-> ~P ^ ~Q Demorgan's law
???????????????

Answer 88

-.- if that's the case, those things were in front of my face.

Answer 89

@eliassaab

Answer 90

~( P V Q) <-> ~P ^ ~Q Demorgan's law
~(P ^ Q) <-> ~P V ~Q Demorgan's Law

These I always remembered as being like a -ve sign in algebra. It "flips" the sign of the inequality (the v becomes ^ and vice versa)

Answer 91

I know that, but it can't be the entire answer right? @wio

Answer 92

It is possible that a tautology can be proven with a single property of Boolean Algebra.

Answer 93

Why would it not be sufficient? Is there some number-of-steps quota you are hoping to reach?

Answer 94

ummm no :/

Answer 95

single property...like 1 line proofs