2.5k views

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. $(\neg p ∧ \neg q) ∨ (p ∧ q)$

edited | 2.5k views

$(p \iff q)$

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

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

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

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

by Junior (803 points)
edited
+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)
+1
There are two options in SOP form. And as biconditional is equivalent XNOR logic the answer is quite straight forward.
0
+5

@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
option c
by Active (1.2k points)