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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

1. True
2. False
3. Same as the truth-value of $b$
4. Same as the truth-value of $d$

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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.

@ leen how did you change c to b?
@sid1221 it is given in question that 'b' is equivalent to 'c'(b$\Leftrightarrow$c)
o yes thanks ..i missunderstood
It is given that a⇔(b∨∼b) which implies a⇔T so there’s nothing wrong if we put a as T right?
Yes, we can use a=T and b=c.

Given that, a$\Leftrightarrow$b∨~b

It is equivalent to a$\Leftrightarrow$TRUE

$\therefore$ (a∧b)$\rightarrow$((a∧c)∨d)

wkt, 1∧x = x

$\therefore$ (a∧b) = 1∧b = b

similarly, 1∧c = c

We now have, b $\rightarrow$(c∨d)

Which can be written as,

~b∨c∨d

We also know that b$\Leftrightarrow$c

$\therefore$ ~b∨c = TRUE

$\therefore$ TRUE∨d = TRUE

And hence answer is option a

Explanation is correct, just to highlight the following statement

We also know that b$\Leftrightarrow$c  holds

$\therefore$ ~b∨c = TRUE

if you mean  b$\Leftrightarrow$c   means   ~b∨c = TRUE   then its not true. Although you may not mean this, but while reading the answer it seems that.

Nice Explanations
great explanation
a ⇔ ( b V ~b) = and a ⇔  True means both a and True are equivalent
b ⇔c means both b and c are equivalent

(a ∧ b) → (a ∧ c) ∨ d
=(True ∧ b) → (True ∧ c) ∨ d     (a ⇔  True)
= b → c ∨ d
= ~b ∨ c ∨ d
= ~b ∨ b ∨ d     (b ⇔c)
= True  ∨ d
=True

Hence ans is A

This will be helpful. see my solution:

Nyc.....
Nice hand writing ...

Here it is given that

a ↔(b v ¬b) ≡ a ↔ True

so value of a will be true

Now b ↔ c means when b is true c will be true true or when b is false c will be false

case 1: Let's take b as TRUE

so, c will be TRUE

so, for expression [(a∧b)→(a∧c)∨d ]

a is true and b  is true so a∧b true and  a∧c true so both LHS and RHS are true hence the expression will be true

when b is false c is also false  as LHS is false the expression will always be true