Proof by Contradiction to Prove Theorem that is not True

Question

Based on the following incorrect Theorem, why is this proof by contradiction not valid?

Theorem
If $x+y=10$
Then $x\neq 3 \text{ and } y\neq 8$

Proof
Assume:
$x+y=10$
$x=3$ or $y=8$

Let $x=3$ and $y=8$
Since $x+y=3+8=11\neq 10$ we have a contradiction based on the assumptions.

Based on this haven't we just proven the above theorem which we know to be incorrect. The steps I have been taught for proving something using proof by contradiction is to assume $P\land \lnot Q$ which I believe that I have done correctly. Then you simply want to derive something that is not true, which I also feel that I have done ($x+y\neq 10$).

So, my question is, where have I gone wrong/what is it that I am not understanding correctly?

Answer 1

Yes, you have set up things correctly: Assume $x+y=10$, and then, to start the proof by Contradiction, you (correctly) assumed that it is not true that $x \neq 3$ and $y \neq 8$. And the latter means that $x=3$ or $y=8$. But from that, you cannot infer that $x=3$ and $y=8$, which is what you need in order to get to $x+y=11$ and thus to your contradiction.

Answer 2

this theorem is wrong

$P: \bigg[x+y=10\implies (x\ne 3\land y\ne 8)\bigg]\iff \bigg[(x=3 \lor y=8)\implies x+y\ne 10\bigg]$

Suppose $x=3$, you take $y:=7$ so $x+y=10$ contradiction so your theorem is wrong

by contradiction :

$\lnot P :\quad x+y=10 \land (x= 3\lor y= 8)$

$x+y=10$ and $x=3$ we take $y:=7$, so the proposition $\lnot P$ is right.

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I remind you $P:(s\implies t)\iff (\lnot s\lor t)$ so $\lnot P :s\land \lnot t$ To avoid any confusion you have to add quantifier so the proposition becomes

$ P :\forall (x,y)\in \mathbb{N}^2,\bigg[x+y=10\implies (x\ne 3\land y\ne 8)\bigg]$

So the negative is :

$\lnot P : \exists (x,y)\in \mathbb{N}^2,\quad (x+y=10) \land (x= 3\lor y= 8)$

We only need to find one 2-tuple that satisfies the proposition, so $(x,y)=(3,7)$ fits. We conclude $\lnot P$ is right thus $P$ is wrong