Let $a, b, c, d$ be propositions. Assume that the equivalence $a ⇔ ( b \vee \neg b)$ and $b ⇔c$ hold. Then the truth-value of the formula $(a ∧ b) → (a ∧ c) ∨ d$ is always
Given that $\ a\Leftrightarrow (b\vee \sim b)$ and $ b\Leftrightarrow c$ Now, $(a\wedge b)\rightarrow (a \wedge c)\vee d$ $\equiv (a\wedge b)\rightarrow (a \wedge b)\vee d$ $(\because b \Leftrightarrow c)$ $\equiv \neg (a\wedge b)\vee (a \wedge b)\vee d$ $\equiv T \vee d$ $\equiv T$ Hence, Option(A) True.
Given that, ab∨~b
It is equivalent to aTRUE
(a∧b)((a∧c)∨d)
wkt, 1∧x = x
(a∧b) = 1∧b = b
similarly, 1∧c = c
We now have, b (c∨d)
Which can be written as,
~b∨c∨d
We also know that bc
~b∨c = TRUE
TRUE∨d = TRUE
And hence answer is option a
Explanation is correct, just to highlight the following statement
We also know that bc holds
if you mean bc means ~b∨c = TRUE then its not true. Although you may not mean this, but while reading the answer it seems that.
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