# Peculiar proof

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

Since $0\leq x\leq 1$ ,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 :

$x+y\leq 2$.

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

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Isn't that rather obvious?

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

Edited by triclino
not complete

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So there are only three cases, not four.

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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.

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

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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).

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

Would you use simplification again??

No. Clearly not.

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

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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.

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

And just how do we do that??

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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.

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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 $|x+y|\leq 2$ ,then : $2<|x+y|\leq 2$ .Which of course results in 2>2

Edited by triclino
correction

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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.

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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 by triclino
correction

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Contradiction is: $q\wedge\neg q$

q here is 1=0 and not q is $1\neq 0$

$1\neq 0$ is a field axiom

(x=0 and x=1) is equivalent to 1=0 ,and not(x=0 and x=1) is equivalent to:$1\neq 0$

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Contradiction is: $q\wedge\neg q$

No. While you are correct that $q\wedge\neg q$ 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.

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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 .

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