Math Logic - Printable Version

Math Logic - Printable Version

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Math Logic - Manu - 2011-03-08

I JUST started with this, and I can't do the exercises, I simply don't know where to start... I have to verify:

Code:
[p^(p=>q)]=>q

What I've got so far is

Code:
~[p^(p=>q)] v q

~p v ~(p=>q) v q

~p v (p ^ ~q) v q

After that step I just don't know what to do, also, just in case my symbols aren't used correctly, ^ conjunction, v as "or" and ~ as "no".
It should be universal language but I hope I didn't do anything stupid just cause I'm asking in english.

Math Logic - OB3LISK - 2011-03-08

I wish I knew about computers because I do know about math logic.

Btw it looks like you have an unnecessary space in that first one.

Math Logic - Shidoshi - 2011-03-08

OB3LISK Wrote:I wish I knew about computers because I do know about math logic.

Btw it looks like you have an unnecessary space in that first one.

This is not at all related to computers or programming.
I can't really help since I never had any math logic classes, this all seems greek to me.

Math Logic - HellenzSin - 2011-03-08

Code:
[p^(p=>q)]=>q

~[p^(p=>q)] v q

~p v ~(p=>q) v q

~p v (p ^ ~q) v q

~p v ~q v q

~p v 1 

1

Math Logic - Kalovale - 2011-03-08

Darkmaniak Wrote:I JUST started with this, and I can't do the exercises, I simply don't know where to start... I have to verify:

Code:
[p^(p=>q)]=>q
What I've got so far is

Code:
~[p^(p=>q)] v q ~p v ~(p=>q) v q ~p v (p ^ ~q) v q

After that step I just don't know what to do, also, just in case my symbols aren't used correctly, ^ conjunction, v as "or" and ~ as "no".
It should be universal language but I hope I didn't do anything stupid just cause I'm asking in english.

1/ Verification: confirmation: additional proof that something that was believed (some fact or hypothesis or theory) is correct. (Just so I don't get off track, I haven't really done any of these exercises)
2/ Symbology:
- ∧ = conjunction: A∧B is satisfied if and only if both A and B are true
- ⇒ = implication: A⇒B means if A is true, then B is true
- ¬ = not: The famous P ∧ ¬P that can derive anything under this or the next sky, including your mom's phone number

[p ∧ (p⇒q)] ⇒ q translates into:
Layer 1: If [p ∧ (p⇒q)] is true, then q is true.
Layer 2: If [p ∧ (p⇒q)] is true, p and (p⇒q) must simultaneously be true.
Layer 3: Well, it's obvious at this point.

Basically, this reads as: "If P is true and if p's trueness leads to q's trueness, then q is true." which is pretty damned annoying if you ask me.
So we flip things around.

If the claim is not true, then:
- p is not true, or
- p implies q is not true
while q is still true.
p implies q being untrue means p and ¬q have to be true at the same time.

q and ¬q can't be true at the same time, so yeah.

Math Logic - Manu - 2011-03-08

HellenzSin Wrote:
Code:
~p v (p ^ ~q) v q ~p v ~q v q

That's the step that intrigues me, how do you get to do that, what properties are you using?

Math Logic - OB3LISK - 2011-03-08

Shidoshi Wrote:This is not at all related to computers or programming.

You're telling me what Kalovale said it's computery?!

Talk in English and I'll be able to help.

Math Logic - Kalovale - 2011-03-08

Darkmaniak Wrote:That's the step that intrigues me, how do you get to do that, what properties are you using?

Logically speaking, I think "p implies q" being untrue means p has to be true but does not point towards q (thus ¬q is true simultaneously with p), because if p is not true and p implies q is true, it doesn't lead anywhere.

Math Logic - HellenzSin - 2011-03-08

Darkmaniak Wrote:That's the step that intrigues me, how do you get to do that, what properties are you using?

Don't remember the name, but:

p ^ (~p v q) = p ^ q
~(p^(~p v q)) = ~(p ^ q)
~p v (p ^ ~q) = ~p v ~q

Math Logic - shouri - 2011-03-08

Darkmaniak Wrote:I JUST started with this, and I can't do the exercises, I simply don't know where to start... I have to verify:

Code:
[p^(p=>q)]=>q
What I've got so far is

Code:
~[p^(p=>q)] v q ~p v ~(p=>q) v q ~p v (p ^ ~q) v q

After that step I just don't know what to do, also, just in case my symbols aren't used correctly, ^ conjunction, v as "or" and ~ as "no".
It should be universal language but I hope I didn't do anything stupid just cause I'm asking in english.

Kinda find hellen's transition from line 4 to 5 a bit iffy. Then again I haven't done this in a while. Found a less iffy way of doing it.

Code:
~[p^(p=>q)] v q

~[p^ (~p v q)] v q

~[ (p^~p) v (p^q)] v q  by distributivity law

~[ 0 v (p^q)] v q

~[ p^q] v q 

(~p v ~q) v q

since these are all ORs, we can use associativity now

~p v (~q v q)

~p v 1  = 1

so the original statement is always true.

Math Logic - Manu - 2011-03-08

HellenzSin Wrote:Don't remember the name, but:

p ^ (~p v q) = p ^ q
~(p^(~p v q)) = ~(p ^ q)
~p v (p ^ ~q) = ~p v ~q

=/ assuming that's a property, we weren't taught that, closest is absorption (p ^ (p v q) = p), do you think there's another way of doing it? otherwise the professor told us to do something we can't do yet...

Edit:
Thanks for that Big Grin

shouri Wrote:~[ (p^~p) v (p^q)] v q by distributivity law
~[ 0 v (p^q)] v q

I left it there and then took the ~ to get to:

1 v ~(p^q) v q
1 v q
1

ok now, here I don't even know what I have to do: (asked to deny.. but what does that even mean)

( $[Image: d4d49bead125261b226eaa867bd016ce.png]$ x $[Image: 8c20c78b364ed5dbadd49e5b997aa1cc.png]$ $[Image: 6e0372ee3e41f8ed7bba429e0ccdc96e.png]$ ) [p(x) ^q(x)] => ( $[Image: 93ebe8636e1f8d60004fe33d1321674e.png]$ x)[p(x) v q(x)]

Math Logic - HellenzSin - 2011-03-08

The above works. Anyways I was thought that lol, but I know they aren't common at all.

Math Logic - Shidoshi - 2011-03-08

Properties are logical (and commonly used) shortcuts so all exercices you do can be said to be a "property"

Math Logic - HellenzSin - 2011-03-08

Woops....