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Which one of the following is NOT equivalent to $p ↔ q$?

  1. $(\neg p ∨ q) ∧ (p ∨ \neg q)$
  2. $(\neg p ∨ q) ∧ (q → p)$
  3. $(\neg p ∧ q) ∨ ( p ∧ \neg q)$
  4. $(\neg p ∧ \neg q) ∨ (p ∧ q)$
in Mathematical Logic by Boss (30.8k points)
edited by | 2.6k views

2 Answers

+33 votes
Best answer
$(p \iff  q)$

$= (p\to q)\wedge (q\to p)$

$= (\neg p\vee q)\wedge (q\to p)$       As$(p\to q = \neg p\vee q )$

$=(\neg p\vee q)\wedge  (\neg q\vee p)$

$=(\neg p\wedge \neg q)\vee (p\wedge q)$

So, answer C
by Junior (803 points)
edited by
+1
how to obtain step 4 of the solution from its step 3 ?
+13
p ↔ q    = (p→ q) ∧ (q→p)

              =(¬p∨q) ∧ (q→p) --------option(B)

              =(¬p ∨ q) ∧ (¬q ∨ p) = (¬p ∨ q) ∧ ( p ∨ ¬q)------------option(A)

               = (¬p ∧ p ) ∨ (¬p ∧ ¬q ) ∨ (q ∧ p ) ∨ (q ∧ ¬q )  (Distributive law)

               = (¬p ∧ ¬q ) ∨ (q ∧ p ) = (¬p ∧ ¬q ) ∨ (p ∧ q )----------option(D)   ((¬p ∧ p )=0,(q ∧ ¬q )=0)
+2
There are two options in SOP form. And as biconditional is equivalent XNOR logic the answer is quite straight forward.
0
Please explain for option d
+6

@Swami patil   (¬p ∨ q) ∧ (¬q ∨ p)(Convert into SOP from)

=(p' + q)(q' + p)

=(p'q' + qp)

 =(¬p ∧ ¬q ) ∨ (p ∧ q ) Option (d)

0
Thanks Brother
+3 votes
option c
by Active (1.2k points)
Answer:

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