j.r-xrs

I am having a major problem with tautologies and logical connectives!

I understand what a tautology is and how to prove some Boolean formulae ie. ones with 2 variables are tautologies, but after that I'm lost.

I understand how to do truth tables for instance:

For the Boolean formulae P and Q state whether or not the formula P ≡ Q is a tautology:

a) P = X → Y, Q = ¬X → ¬Y
Which I believe to read “P equals X implies Y, Q equals not X implies not Y”

Therefore to solve when I have worked out all possible values of P and Q I can compare P and Q to see if they are equivalent (P ≡ Q) :



This example would not be a tautology due to lines 2 and 3 of the table where P is not equivalent to Q.

b) P = X ٨ (Y ٧ Z), Q = X ٧ (X ٨ Z)
Which I think reads: "P equals X and (Y or Z), Q equals X or (X and Z)"

Can anyone (is it possible to?) draw up a truth table like the one above and tell me whether it is a tautology or not?

Much appreciated.

apples24

iTrader: (1)

**** me ya lost me on the first line lol

j.r-xrs

Yeah sorry, it's always hard to explain something you dont understand to someone else and I guess it's even harder for them to tell you the right answer after your crap explanation! lol

carl

Isn't "implies that" a double arrow, kind of like =>

Anyway, your original equations for P and Q are the same (you have no "not X" and "not Y" symbols)

I'm not sure how you get that if X is F is Y is T then X=>Y is T (third line down).

It's been a while since I did this stuff.

Surel for (b) you expand it out?

j.r-xrs

How do you do that?

And

If P ≡ Q does this equate to:

(P → Q) ٨ (Q → P)

Or is that total rubbish?

Dazza01

Lost me on the subject line, just had to have a nosey to see if i knew what the hell it was about

carl

Well the first thing you do in (b) is get rid of the brackets.

P = X (Y + Z) = XY + YZ
Q = X + (XZ) = X + XZ

Does that help? I'm used to using the . or multiplication symbol for AND and the + for OR.

Might find some help here: http://www.ee.surrey.ac.uk/Projects/...w/boolalgebra/
PS: in that link a letter with a hat is NOT.

XRS

Have you seen/followed this?

http://people.bath.ac.uk/ma0jbm/Tautologies.htm

Or is this chap the source of your problem?

ALi-B

iTrader: (1)

**** me, I remember doing this......

don't mind if I run away screaming will you?

j.r-xrs

top

speedking

but am prepared to have a go

line X Y Z (Y or Z) P (X and Z) Q
1 T T T T T T T
2 T T F T T F T
3 T F T T T T T
4 T F F F F F T
5 F T T T F F F
6 F T F T F F F
7 F F T T F F F
8 F F F F F F F

not a tautology because of line 4. Apologies for the scoobynet auto formatting There are 8 columns.

Found this but don't know how useful or accurate it will be.

HTH

speedking

PS
http://en.wikipedia.org/wiki/Truth_table

XRS

Deserves a BTT for getting up at 7 in the morning

j.r-xrs

speedking,

Thats awesome, thanks very much.

That program is a nice way of checking things through as well

Really appreciate it

speedking

I hope you get your GCSE

speedking

BTW

if A is (P → Q) ٨ (Q → P)

P Q P=>Q Q=>P A
T T T T T
T F T F F
F T F T F
F F T F F

A=P^Q

Sprint Chief

Are you sure the last line of your truth table is right on post #16 Speedking

Conclusion looks a bit shaky too...

Pay attention at the back!