Prove that $ p \Leftrightarrow q = (p \Rightarrow q) \cdot (q \Rightarrow p) $.

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To prove this equivalence, a truth table is constructed to compare the truth values of the bi-conditional expression $ p \Leftrightarrow q $ with the result of the expression $ (p \Rightarrow q) \cdot (q \Rightarrow p) $ for all possible truth value combinations of $ p $ and $ q $.

    Truth Table for Bi-conditional Decomposition:
    
        
            
                $ p $
                $ q $
                $ p \Leftrightarrow q $
                $ p \Rightarrow q $
                $ q \Rightarrow p $
                $ (p \Rightarrow q) \cdot (q \Rightarrow p) $
            
        
        
            
                0
                0
                1
                1
                1
                1
            
            
                0
                1
                0
                1
                0
                0
            
            
                1
                0
                0
                0
                1
                0
            
            
                1
                1
                1
                1
                1
                1
            
        
    

    Conclusion:
    Since the columns for $ p \Leftrightarrow q $ and $ (p \Rightarrow q) \cdot (q \Rightarrow p) $ are identical, they possess the same truth set: (1, 0, 0, 1).
    Therefore, it is proved that: $ p \Leftrightarrow q = (p \Rightarrow q) \cdot (q \Rightarrow p) $.

\( p \)	\( q \)	\( p \Leftrightarrow q \)	\( p \Rightarrow q \)	\( q \Rightarrow p \)	\( (p \Rightarrow q) \cdot (q \Rightarrow p) \)
0	0	1	1	1	1
0	1	0	1	0	0
1	0	0	0	1	0
1	1	1	1	1	1

Prove that \( p \Leftrightarrow q = (p \Rightarrow q) \cdot (q \Rightarrow p) \).

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