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]??

Isn't that rather obvious?

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

Edited May 21, 201015 yr by triclino
not complete

So there are only three cases, not four.

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.

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

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).

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??

No. Clearly not.

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

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.

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

 

And just how do we do that??

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.

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, 201015 yr by triclino
correction

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.

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, 201015 yr by triclino
correction

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

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]

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.

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 .