Question

Let a, b, c , d be propositions. Assume that the equivalence $a\leftrightarrow (b\vee \rightharpoondown b)$ and b $\leftrightarrow$ c hold. Then the truth - value of the formula $(a\wedge b)\to (a\wedge c)\vee d$ is always

• A. True
• B. False
• C. Same as the truth-value of b
• D. Same as the truth-value of b

Answer 1

The correct option is A True

(a)
$a\leftrightarrow (b\vee \rightharpoondown b)$
$a\leftrightarrow$ True
So a is True , i.e. a = 1

$b\leftrightarrow$ c holds. So b = c
Now the given expression is $(a\wedge b)\to ((a\wedge c)\vee b)\equiv (a\cdot b)\to ((a\cdot c)+d)$
Putting a = 1 in above expression we get
$1\cdot b\to ((1\cdot c)+d)$
$\equiv b\to c+d$
$\equiv b{}^{\prime }+c+d$

Now putting b = c is above expression we get
$c{}^{\prime }+c+d\equiv 1+d\equiv 1$

So the expression is always true.