Peculiar proof

[triclino]

In proving : [math] x+y\leq 2[/math] given that : [math]0\leq x\leq 1[/math] and y=1,the following proof is pursued:

 

Since [math]0\leq x\leq 1[/math] ,then (x>0 or x=0)and (x<1 or x=1) ,which according to logic is equivalent to:

 

(x>0 and x<1) or (x>0 and x=1) or (x=0 and x<1) or ( x=0 and x=1).

 

And in examining the three cases ( except the 4th one : x=0 and x=1) we end up with :

 

[math] x+y\leq 2[/math].

 

The question now is : how do we examine the 4th case so to end up with [math] x+y\leq 2[/math]??

[D H]

Isn't that rather obvious?

[triclino]

No, not at all because x=0 and x=1 results in 0=1

Edited May 21, 2010 by triclino
not complete

[shyvera]

So there are only three cases, not four.

[the tree]

It seems a rather convoluted way to go about it.

 

( (x>0) and (x<1) )

 

=>

 

(x < 1)

 

=>

 

(x < 1 ) or ( x = 1)

 

Only those two cases need to be looked at, neither of them being degenerate as in your example.

[triclino]

Your kind of proof is based on what fact(s)?

[the tree]

I didn't present a proof, I just presented the cases required, but conjunction elimination anyway. (apparently more commonly known as simplification, but I hate that term).

[triclino]

And if the problem asked ,instead of the statement :[math]x+y\leq 2[/math],the statement : [math]|x+y|\leq 2[/math] to be proved??

 

Would you use simplification again??

[the tree]

No. Clearly not.

[triclino]

Then how do we treat the 4th case in this new problem??

[the tree]

I'm tempted to say cast it aside as degenerate. More formally, demonstrate that the first three cases are the only ones that can actually exist.

[triclino]

. More formally, demonstrate that the first three cases are the only ones that can actually exist.

 

And just how do we do that??

[the tree]

Roughly:

 

Taking not(x=0 and x=1) as given,

 

( (x>0 and x=1) or ( x=0 and x=1) ) and not(x=0 and x=1)

=>

(x>0 and x=1)

 

by disjunctive syllogism.

[triclino]

Roughly:

 

Taking not(x=0 and x=1) as given,

 

( (x>0 and x=1) or ( x=0 and x=1) ) and not(x=0 and x=1)

=>

(x>0 and x=1)

 

by disjunctive syllogism.

 

On the other hand i could use disjunctive syllogism and end up with : 2>2

 

Since x=0 and x=1 => 0=1 => (0=1) or |x+y|>2 and not (0=1) => |x+y|>2.

 

But since [math]|x+y|\leq 2[/math] ,then : [math]2<|x+y|\leq 2[/math] .Which of course results in 2>2

Edited May 22, 2010 by triclino
correction

[D H]

No you can't, triclino. x=0 and x=1 implies absolutely nothing. x=0 and x=1 is a contradiction. The only conclusion that can drawn from (1) P -> Q and (2) ~P is that P is false.

[triclino]

No you can't, triclino. x=0 and x=1 implies absolutely nothing. .

 

On the contrary D.H contradiction can imply everything.

 

Thru disjunction syllogism as Tree pointed out you can imply everything.

 

Learn disjunction syllogism.

 

The contradiction here is: (x=0 and x=1) and not( x=0 and x=1)

Edited May 25, 2010 by triclino
correction

[D H]

The contradiction here is x=0 and x=1. Think about it.

[triclino]

The contradiction here is x=0 and x=1. Think about it.

 

 

Contradiction is: [math]q\wedge\neg q[/math]

 

q here is 1=0 and not q is [math]1\neq 0[/math]

 

You thing twice about it

 

[math]1\neq 0[/math] is a field axiom

 

(x=0 and x=1) is equivalent to 1=0 ,and not(x=0 and x=1) is equivalent to:[math]1\neq 0[/math]

[D H]

Contradiction is: [math]q\wedge\neg q[/math]

No. While you are correct that [math]q\wedge\neg q[/math] is a contradiction, it is not the only thing that qualifies as a contradiction. A contradiction is any statement that is unsatisfiable.

 

x=0 and x=1 is unsatisfiable. It is a contradiction.

[triclino]

A contradiction is any statement that is unsatisfiable.

 

x=0 and x=1 is unsatisfiable. It is a contradiction.

 

 

You are entangling False with contradiction .

 

x=0 and x=1 is simply a false statement.

 

This is basic stuff in Symbolic Logic, you better learn about it .