MHB Logic problem with arithmetic and inequalities

AI Thread Summary
The discussion centers around a logic problem involving arithmetic and inequalities, specifically the invalid conclusion that "2 is less than or equal to zero." Participants analyze the premises using logical statements and transformations, highlighting the flaws in reasoning. They emphasize that the initial statements can be manipulated but do not lead to the conclusion that 2 is less than or equal to zero. The conversation also touches on the implications of defining logical statements and the necessity of sound reasoning in formal systems. Ultimately, the conclusion cannot be proven within a sound logical framework.
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Given:

1)it is not true that : 2>0 and 2+3 =7

2)if it is not true that 2>0 then 2 is less or equal to zero

3)if 2+3 =7 ,then 3+3 =8

4) but 3+3 is not equal to 8

Then prove:

2 is less or equal to zero
 
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Re: logic

It's an invalid conclusion, obviously. The problem comes as follows:

Let $A$ be the statement that $2>0$, and $B$ be the statement that $2+3=7$. Let $C$ be the statement that $2\le 0$. Let $D$ be the statement that $3+3=8$. Then your premisses are as follows:

\begin{align*}
& \lnot(A \land B) \\
& \lnot A \implies C \\
& B \implies D\\
& \lnot D\\
& \therefore C.
\end{align*}
The first statement can be transformed, via DeMorgan, to
$$\lnot A \lor \lnot B.$$
So your assumption of $\lnot D$ could, via modus tollens, give you $\lnot B$. But then, analyzing the first statement in its DeMorgan form, you are now stating that one of the options of the disjunction is true. That in no way implies that the other disjunct is true. So your reasoning chain ends. You cannot claim that $\lnot A$ is true.
 
Last edited:
Re: logic

Ackbach said:
It's an invalid conclusion, obviously. The problem comes as follows:

Let $A$ be the statement that $2<0$, and $B$ be the statement that $2+3=7$. Let $C$ be the statement that $2\le 0$. Let $D$ be the statement that $3+3=8$. Then your premisses are as follows:

\begin{align*}
& \lnot(A \land B) \\
& \lnot A \implies C \\
& B \implies D\\
& \lnot D\\
& \therefore C.
\end{align*}
The first statement can be transformed, via DeMorgan, to
$$\lnot A \lor \lnot B.$$
So your assumption of $\lnot D$ could, via modus tollens, give you $\lnot B$. But then, analyzing the first statement in its DeMorgan form, you are now stating that one of the options of the disjunction is true. That in no way implies that the other disjunct is true. So your reasoning chain ends. You cannot claim that $\lnot A$ is true.

why should you not put :

$\neg B$ for $3+2=7$ since $2+3=7$ is false
 
Re: logic

solakis said:
why should you not put :

$\neg B$ for $3+2=7$ since $2+3=7$ is false

Evgeny can correct me if I'm wrong, but I think if you're in a two-valued logic system, where $\lnot( \lnot B)=B$, then it doesn't matter which you use - just a matter of definition. If you choose $B$ the way I have, it's a false proposition. If you choose your definition, it's a true proposition. You'd have to change your assumptions if you changed your definition, but the logic would work out analogously.
 
Re: logic

Ackbach said:
Let $A$ be the statement that $2<0$
This should say, $2 > 0$.

solakis said:
why should you not put :

$\neg B$ for $3+2=7$ since $2+3=7$ is false
One has the right to introduce any notation. Abbreviating some expression by a variable is not a logical step; it does not change a problem in any essential way,

The premises in the OP are true, say, on integers, and the conclusion is not. So the conclusion cannot be proved in any formal system that is sound with respect to integers. (Regular logic is sound with respect to all models.)
 
Re: logic

Evgeny.Makarov said:
This should say, $2 > 0$.

Thank you! I've corrected that.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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