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Prove: If $x+y>22$ then $x>11$ and $ y>11$

I'm not sure but I believe you have to use proof by contradiction.

EDIT: Proof: By contradiction, let $x+y>22$ and $x \leq 11$ or $y \leq 11$. Adding both sides, we get $x+y \leq 22$ which is a contradiction since we said $x+y>22$.

Any corrections?

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    $\begingroup$ What about $x=3$, $y=20$? Or do you want an"or" instead of "and?" $\endgroup$
    – Paul
    Mar 2 '16 at 21:42
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    $\begingroup$ You mean $x>11$ or $y>11$? $\endgroup$
    – k99731
    Mar 2 '16 at 21:43
  • $\begingroup$ Did you mean $x\gt11$ OR $y\gt11$ $\endgroup$
    – Mufasa
    Mar 2 '16 at 21:43
  • $\begingroup$ ok let me edit. $\endgroup$
    – kero
    Mar 2 '16 at 21:44
  • $\begingroup$ As per Paul's comment, your proof must be wrong, due to the fact that the conditional in the title and in the first line of the question is false. $\endgroup$ Mar 2 '16 at 21:46
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You cannot add two equations joined by a logical or. The original statement is false so there is nothing to prove. If you replace the and in the original statement with or, then the or in your proof becomes an and, making the proof valid.

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  • $\begingroup$ What do you mean by the original statement is false? $\endgroup$
    – kero
    Mar 2 '16 at 21:50
  • $\begingroup$ I mean that the statement "if $x+y>22$ then $x>11$ and $y>11$" is logically false, as has been amply illustrated by commenters. $\endgroup$ Mar 4 '16 at 12:31
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Ok, if $x<11$ AND $y<11$ you can sum it into $x+y<22$, but if its OR you have a complex, not a system, you can't sum it(even it would be eqs). For example, x=13 or y=2 doesn't mean x+y=15.

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