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Logic problem with arithmetic and inequalities

solakis

Active member
Dec 9, 2012
321
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
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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:

solakis

Active member
Dec 9, 2012
321
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.
why should you not put :

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

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: logic

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.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Re: logic

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

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.)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197