in Mathematical Logic edited by
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40 votes
40 votes

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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6 Answers

43 votes
43 votes
Best answer

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.

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4 Comments

o yes thanks ..i missunderstood
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It is given that a⇔(b∨∼b) which implies a⇔T so there’s nothing wrong if we put a as T right?
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Yes, we can use a=T and b=c.
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49 votes
49 votes

Given that, a\Leftrightarrowb∨~b

It is equivalent to a\LeftrightarrowTRUE

\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\Leftrightarrowc

\therefore ~b∨c = TRUE

\therefore TRUE∨d = TRUE

And hence answer is option a

3 Comments

Explanation is correct, just to highlight the following statement

We also know that b\Leftrightarrowc  holds

\therefore ~b∨c = TRUE

if you mean  b\Leftrightarrowc   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
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great explanation
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12 votes
12 votes
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
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6 votes
6 votes

This will be helpful. see my solution: 

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2 Comments

Nyc.....
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Nice hand writing ...
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Answer:

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