Math Logic

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.

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.

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.

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

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.

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?

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.

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.

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

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.

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

~[ (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)

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

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

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

Woops....