Can Someone solve this question please
(B-->R) and (BL --> Y) and (BL --> ~ R) --> (B --> Y)
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.
You might need more paretheses. Is it:
$\displaystyle ((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow(B\rightarrow Y)$?
Or
$\displaystyle (B\rightarrow R)\wedge (BL\rightarrow Y)\wedge ((BL\rightarrow {\sim}R)\rightarrow(B\rightarrow Y))$?
Anyway, you might start by breaking down each part. $\displaystyle B\rightarrow R \equiv {\sim}B\vee R$.
$\displaystyle BL\rightarrow Y \equiv {\sim}BL\vee Y$
$\displaystyle BL\rightarrow {\sim}R \equiv {\sim}BL \vee {\sim}R \equiv {\sim}(BL\wedge R)$
$\displaystyle B\rightarrow Y \equiv {\sim}B \vee Y$
Does that help at all?
$\displaystyle \begin{align*}((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R)) & \equiv (({\sim}B\vee R)\wedge ({\sim}BL\vee Y)\wedge ({\sim}BL\vee {\sim}R)) \\ & \equiv (({\sim}B\vee R)\wedge ({\sim}BL \vee (Y\wedge {\sim}R))) \\ & \equiv (({\sim}B\wedge {\sim}BL) \vee ({\sim}B\wedge Y \wedge {\sim}R) \vee (R\wedge {\sim}BL))\end{align*}$
Next, we negate that statement:
$\displaystyle \begin{align*}{\sim}(({\sim}B\wedge {\sim}BL) \vee ({\sim}B\wedge Y \wedge {\sim}R) \vee (R\wedge {\sim}BL)) & \equiv ((B\vee BL)\wedge (B\vee {\sim}Y\vee R)\wedge ({\sim}R\vee BL)) \\ & \equiv ((B\wedge {\sim}R) \vee (B\wedge BL) \vee (BL\wedge {\sim}Y) \vee (BL\wedge R))\end{align*}$
Finally, putting it all together:
$\displaystyle \begin{align*}((B\rightarrow R)\wedge (BL\rightarrow Y)\wedge (BL\rightarrow {\sim}R))\rightarrow (B\rightarrow Y) & \equiv (((B\wedge {\sim}R) \vee (B\wedge BL) \vee (BL\wedge {\sim}Y) \vee (BL\wedge R))\vee ({\sim}B\vee Y)) \\ & \equiv {\sim}(B\wedge R\wedge {\sim}BL) \\ & \equiv {\sim}B \vee {\sim}R \vee BL \\ & \equiv (B\wedge R)\rightarrow BL\end{align*}$
So, the whole thing simplifies to $\displaystyle (B\wedge R)\rightarrow BL$. I may have made a mistake. I typed this up quickly. So, it is not valid if that is false.
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?
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.