# Thread: Solve without the use of truth tables

1. ## Solve without the use of truth tables

Can Someone solve this question please

(B-->R) and (BL --> Y) and (BL --> ~ R) --> (B --> Y)

2. ## Re: Solve without the use of truth tables

What do you mean by solving this question? This formula is not a tautology.

3. ## Re: Solve without the use of truth tables

Originally Posted by emakarov
What do you mean by solving this question? This formula is not a tautology.
I meant determine the validity of the argument without the use of truth tables.Show how it is invalid

4. ## Re: Solve without the use of truth tables

There are well-known methods, other than truth tables, for proving that a formula is a tautology (though none of these methods is "standard"), but methods for proving that a formula is not a tautology are not nearly as well-known. Perhaps you have studied one; then you should tell us how it was done in your course.

5. ## Re: Solve without the use of truth tables

You might need more paretheses. Is it:

$((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow(B\rightarrow Y)$?

Or

$(B\rightarrow R)\wedge (BL\rightarrow Y)\wedge ((BL\rightarrow {\sim}R)\rightarrow(B\rightarrow Y))$?

Anyway, you might start by breaking down each part. $B\rightarrow R \equiv {\sim}B\vee R$.

$BL\rightarrow Y \equiv {\sim}BL\vee Y$

$BL\rightarrow {\sim}R \equiv {\sim}BL \vee {\sim}R \equiv {\sim}(BL\wedge R)$

$B\rightarrow Y \equiv {\sim}B \vee Y$

Does that help at all?

6. ## Re: Solve without the use of truth tables

Originally Posted by SlipEternal
You might need more paretheses. Is it:

$((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow(B\rightarrow Y)$?

Or

$(B\rightarrow R)\wedge (BL\rightarrow Y)\wedge ((BL\rightarrow {\sim}R)\rightarrow(B\rightarrow Y))$?

Anyway, you might start by breaking down each part. $B\rightarrow R \equiv {\sim}B\vee R$.

$BL\rightarrow Y \equiv {\sim}BL\vee Y$

$BL\rightarrow {\sim}R \equiv {\sim}BL \vee {\sim}R \equiv {\sim}(BL\wedge R)$

$B\rightarrow Y \equiv {\sim}B \vee Y$

Does that help at all?
yes it helps a bit.But is the initial statement valid?

$((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow(B\rightarrow Y)$?

7. ## Re: Solve without the use of truth tables

\begin{align*}((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R)) & \equiv (({\sim}B\vee R)\wedge ({\sim}BL\vee Y)\wedge ({\sim}BL\vee {\sim}R)) \\ & \equiv (({\sim}B\vee R)\wedge ({\sim}BL \vee (Y\wedge {\sim}R))) \\ & \equiv (({\sim}B\wedge {\sim}BL) \vee ({\sim}B\wedge Y \wedge {\sim}R) \vee (R\wedge {\sim}BL))\end{align*}

Next, we negate that statement:

\begin{align*}{\sim}(({\sim}B\wedge {\sim}BL) \vee ({\sim}B\wedge Y \wedge {\sim}R) \vee (R\wedge {\sim}BL)) & \equiv ((B\vee BL)\wedge (B\vee {\sim}Y\vee R)\wedge ({\sim}R\vee BL)) \\ & \equiv ((B\wedge {\sim}R) \vee (B\wedge BL) \vee (BL\wedge {\sim}Y) \vee (BL\wedge R))\end{align*}

Finally, putting it all together:

\begin{align*}((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow (B\rightarrow Y) & \equiv (((B\wedge {\sim}R) \vee (B\wedge BL) \vee (BL\wedge {\sim}Y) \vee (BL\wedge R))\vee ({\sim}B\vee Y)) \\ & \equiv {\sim}(B\wedge R\wedge {\sim}BL) \\ & \equiv {\sim}B \vee {\sim}R \vee BL \\ & \equiv (B\wedge R)\rightarrow BL\end{align*}

So, the whole thing simplifies to $(B\wedge R)\rightarrow BL$. I may have made a mistake. I typed this up quickly. So, it is not valid if that is false.

8. ## Re: Solve without the use of truth tables

I got almost the same result $(\neg Y\land B\land R)\to BL$, or $Y\lor\neg B\lor\neg R\lor BL$. (If $Y=B=R=\text{True}$ and $BL=\text{False}$, then $(B\land R)\to BL$ is false, but the original formula is true.) However, how do we know that $Y\lor\neg B\lor\neg R\lor BL$ is not a tautology? It is obvious, of course, but mostly because it is clear which truth values make this formula false. Then why not use truth values with the original formula?

9. ## Re: Solve without the use of truth tables

Thanks, I probably had a typo at some point, and carried it forward.

10. ## Re: Solve without the use of truth tables

Oh, wow, Wolframalpha makes that a snap: link

(It wouldn't let me use BL, so I used L.)

11. ## Re: Solve without the use of truth tables

thanks all for the responses. Can this same question be solved using the logical equivalences and rules of inference?
(B-->R) and (BL --> Y) and (BL --> ~ R) --> (B --> Y

12. ## Re: Solve without the use of truth tables

Originally Posted by SlipEternal
Oh, wow, Wolframalpha makes that a snap
That's how I did it.

Originally Posted by mathman11
Can this same question be solved using the logical equivalences and rules of inference?
SlipEternal did use logical equivalences. Concerning rules of inference, most calculi are designed to prove, not disprove, formulas. Since this formula is not a tautology, it cannot be proved. There are some systems (e.g., tableau calculus) that can check formulas for validity. If you are using such calculus, you need to say so. So far you have said nothing about the equivalences and inference rules you are using.

13. ## Re: Solve without the use of truth tables

Originally Posted by emakarov
That's how I did it.

SlipEternal did use logical equivalences. Concerning rules of inference, most calculi are designed to prove, not disprove, formulas. Since this formula is not a tautology, it cannot be proved. There are some systems (e.g., tableau calculus) that can check formulas for validity. If you are using such calculus, you need to say so. So far you have said nothing about the equivalences and inference rules you are using.
the tutor did not give much info on the question. The instructions given were to determine the validity of the argument without the use to truth tables. whats the simplest way to state that its not valid. Suck as a specific statement. Thanks!

14. ## Re: Solve without the use of truth tables

My best guess is to follow SlipEternal's example and to convert the given formula to $Y\lor\neg B\lor\neg R\lor BL$. Since it is obvious that it is not valid, neither is the original formula.

15. ## Re: Solve without the use of truth tables

Originally Posted by emakarov
My best guess is to follow SlipEternal's example and to convert the given formula to $Y\lor\neg B\lor\neg R\lor BL$. Since it is obvious that it is not valid, neither is the original formula.
kool thanks