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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 $d$
edited | 2.1k views

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.

edited by
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@ leen how did you change c to b?
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@sid1221 it is given in question that 'b' is equivalent to 'c'(b$\Leftrightarrow$c)
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o yes thanks ..i missunderstood

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

+3

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.

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Nice Explanations
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
edited

This will be helpful. see my solution:

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

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