Question

Prove: \[\huge (p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)\]
without using truth tables

Answer 1

Is that not the definition of equivalence?

Answer 2

it is...but how do you prove it?

Answer 3

You don't prove definitions...

Answer 4

anything's possible...

Answer 5

and this one is

Answer 6

You have a proof?

Answer 7

Well, all right, suppose
p↔q
Then by definition
(p∧q)∨~(p∨q)
then
(p∨q)→(p∧q)
Suppose p
then
p∨q, by addition
then
p∧q by modus ponens
then
q by simplification
Therefore,
p→q.
Suppose q
Then
q∨p, by addition
p∨q, by commutativity
then
p∧q by modus ponens
then
p, by simplification
Therefore,
q→p
Thus
(p→q)∧(q→p)
Hence,
(p↔q)→(p→q)∧(q→p)

Answer 8

...i suppose one really shouldn't prove definitions......

Answer 9

Actually, this is one of two. The other definition, is the one I used, the one that uses ~,∧, and ∨, instead of →

Answer 10

\[\begin{array}{|c|c|c|c|c|}\hline p&q&p\Rightarrow q&p\Leftarrow q&(p\Rightarrow q) \wedge (p\Leftarrow q)&p\Leftrightarrow q
\\\hline\top&\top&\top&\top&\top&\top
\\\top&\bot&\bot&\top&\bot&\bot
\\\bot&\top&\top&\bot&\bot&\bot
\\\bot&\bot&\top&\top&\top&\top
\\\hline\end{array}\]