If everyone else gets to ask for math help...

[notfred]

so do I, so help me with this problem

Give a formal proof of the following tautology:
(A implies B) implies ((B implies C) implies (A or B implies C))

I haven't even tried it yet, thought I'd just see if anyone else wanted to do my homework for me

 

[AgaBoogaBoo]

I haven't even tried it yet, thought I'd just see if anyone else wanted to do my homework for me

lol

 

[notfred]

No one? Prior's thread has 10 replies. I feel so unloved

 

[Koing]

Originally posted by: notfred
No one? Prior's thread has 10 replies. I feel so unloved

I thought you were going to bitch about people who asked for Maths help!

Well I think I did something like this before in my 1st year. Can't be assed to find it but then it would probably be wrong as my maths is cack

Koing

 

[Shujaa]

You shoulda asked your uni mates to help ya then

 

[gopunk]

you don't sound like you need help otherwise i would

 

[sygyzy]

It would only be appropriate if someone offered to help you, for a fee.

 

[Spencer278]

(A implies B) implies ((B implies C) implies (A or B implies C))

I don't get what the question is but isn't that last fact kind of usless give B implies C

 

[Azraele]

Looking at that makes my head hurt just thinking about it.

BTW, do your own work. At least try it before you attempt to get people to do it for you.

 

[PowerMacG5]

Originally posted by: Spencer278
(A implies B) implies ((B implies C) implies (A or B implies C))

I don't get what the question is but isn't that last fact kind of usless give B implies C

Do you even know what a tautology is?

 

[Kyteland]

Lets hope the formatting isn't screwed up.

(A implies B) implies ((B implies C) implies (A or B implies C))


A | B | C | A->B | B->C | AvB | (AvB)->C | (B->C)->((AvB)->C) | (A->B)->((B->C)->((AvB)->C)
-------------------------------------------------------------------------------------------
T | T | T | T | T | T | T | T | T
T | T | F | T | F | T | F | T | T
T | F | T | F | T | T | T | T | T
T | F | F | F | T | T | F | F | T
F | T | T | T | T | T | T | T | T
F | T | F | T | F | T | F | T | T
F | F | T | T | T | F | T | T | T
F | F | F | T | T | F | T | T | T

Edit: Yup, screwed up. It looks perfect in NOTEPAD

 

[fs5]

Originally posted by: notfred
so do I, so help me with this problem

Give a formal proof of the following tautology:
(A implies B) implies ((B implies C) implies (A or B implies C))

I haven't even tried it yet, thought I'd just see if anyone else wanted to do my homework for me

man 2 years since my last class on that... couldn't help ya here... barely remember the hypo and disjunctive syllogism law.s

 

[newbiepcuser]

Originally posted by: Kyteland
Lets hope the formatting isn't screwed up.

(A implies B) implies ((B implies C) implies (A or B implies C))


A | B | C | A->B | B->C | AvB | (AvB)->C | (B->C)->((AvB)->C) | (A->B)->((B->C)->((AvB)->C)
-------------------------------------------------------------------------------------------
T | T | T | T | T | T | T | T | T
T | T | F | T | F | T | F | T | T
T | F | T | F | T | T | T | T | T
T | F | F | F | T | T | F | F | T
F | T | T | T | T | T | T | T | T
F | T | F | T | F | T | F | T | T
F | F | T | T | T | F | T | T | T
F | F | F | T | T | F | T | T | T

Edit: Yup, screwed up. It looks perfect in NOTEPAD

Truth table, is that what you wanted?

 

[fs5]

Originally posted by: Kyteland
Lets hope the formatting isn't screwed up.

(A implies B) implies ((B implies C) implies (A or B implies C))


A | B | C | A->B | B->C | AvB | (AvB)->C | (B->C)->((AvB)->C) | (A->B)->((B->C)->((AvB)->C)
-------------------------------------------------------------------------------------------
T | T | T | T | T | T | T | T | T
T | T | F | T | F | T | F | T | T
T | F | T | F | T | T | T | T | T
T | F | F | F | T | T | F | F | T
F | T | T | T | T | T | T | T | T
F | T | F | T | F | T | F | T | T
F | F | T | T | T | F | T | T | T
F | F | F | T | T | F | T | T | T

Edit: Yup, screwed up. It looks perfect in NOTEPAD

I think he wanted something more formal than a truth table.

 

[Kyteland]

Like this?

(A->B)->((B->C)->((AvB)->C)
!(A->B)v((B->C)->((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v((B->C)->((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v(!(B->C)v((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v(!(B->C)v(!(AvB)vC) : by the rule x->y == !xvy
!(!AvB)v(!(!BvC)v(!(AvB)vC) : by the rule x->y == !xvy
(A^!B)v(!(!BvC)v(!(AvB)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v((B^!C)v(!(AvB)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v((B^!C)v((!A^!B)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v(B^!C)v(!A^!B)v(C) : Simplified parenthesis
(!B)v(B^!C)v(C) : by the rule (x^y)v(!x^y) == y
!!(!BvC)v(B^!C) : by the rule !!x == x
!(B^!C)v(B^!C) : by the rule !(xvy) == (!x^!y)
T : by the rule !xvx == T

Edit: And what the hell do you mean by "more formal than a truth table"? Truth tables are a valid method of proof. His question never stated to prove it only using logical axioms or equivalences.

 

[fs5]

Originally posted by: Kyteland
Like this?

(A->B)->((B->C)->((AvB)->C)
!(A->B)v((B->C)->((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v((B->C)->((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v(!(B->C)v((AvB)->C) : by the rule x->y == !xvy
!(!AvB)v(!(B->C)v(!(AvB)vC) : by the rule x->y == !xvy
!(!AvB)v(!(!BvC)v(!(AvB)vC) : by the rule x->y == !xvy
(A^!B)v(!(!BvC)v(!(AvB)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v((B^!C)v(!(AvB)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v((B^!C)v((!A^!B)vC) : by the rule !(xvy) == (!x^!y)
(A^!B)v(B^!C)v(!A^!B)v(C) : Simplified parenthesis
(!B)v(B^!C)v(C) : by the rule (x^y)v(!x^y) == y
!!(!BvC)v(B^!C) : by the rule !!x == x
!(B^!C)v(B^!C) : by the rule !(xvy) == (!x^!y)
T : by the rule !xvx == T

Edit: And what the hell do you mean by "more formal than a truth table"? Truth tables are a valid method of proof. His question never stated to prove it only using logical axioms or equivalences.

truth tables are a valid method but many professors (at this stage in the game) won't accept them as a proof.