Propositional calculus [closed]

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Why ( P ^ Q ) === ( ~P -> Q ) ?

How to prove?

Thanks!

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No, it's not: ~P -> Q === P v Q, it is not equivalent to P ^ Q

One way to prove it is to use a truth table:

P | Q | P v Q | ~P | ~P -> Q
0   0     0     1       0        
0   1     1     1       1
1   0     1     0       1
1   1     1     1       1
          ^             ^
          +-------------+
       these are equivalent

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Just look at the truth tables for each expression:

p | q | p v q
--+---+------
T | T | T
T | F | T
F | T | T
F | F | F

p | q | ~p -> q
--+---+--------
T | T | T
T | F | T
F | T | T
F | F | F