Q isn't a condition for P, it is the result of P. That is the difference.
If P => Q and Q => P you have a circular argument. Either both are true from the outset, or both are false from the outset. You can't have P (false) => Q (true) as you could before, because if Q is true, P is true (since you claim Q is a neccessary condition of P).
Going back to your example, you are confusing things yet again. You said "Q is a neccessary condition to affirm P", yet your example to show this is and example of P => Q, not Q => P. You said you can't say that if you throw the ball, it didn't fall (an example of P => Q).
However, you can easily have a ball falling (Q) without P. For instance, if R is dropping a ball. R => Q.
So, we can demonstrate a logical contradiction:
P = Throwing a ball.
R = Dropping a ball.
Q = The ball falls.
P => Q
Q => P
R => Q
P is false (we're not throwing the ball). If R then Q (this is true since R => Q). If Q then P (taken from Q => P). Yet P is false and Q is true, and true => false isn't a case of Q => P.
Q.E.D
If P => Q and Q => P you have a circular argument. Either both are true from the outset, or both are false from the outset. You can't have P (false) => Q (true) as you could before, because if Q is true, P is true (since you claim Q is a neccessary condition of P).
Going back to your example, you are confusing things yet again. You said "Q is a neccessary condition to affirm P", yet your example to show this is and example of P => Q, not Q => P. You said you can't say that if you throw the ball, it didn't fall (an example of P => Q).
However, you can easily have a ball falling (Q) without P. For instance, if R is dropping a ball. R => Q.
So, we can demonstrate a logical contradiction:
P = Throwing a ball.
R = Dropping a ball.
Q = The ball falls.
P => Q
Q => P
R => Q
P is false (we're not throwing the ball). If R then Q (this is true since R => Q). If Q then P (taken from Q => P). Yet P is false and Q is true, and true => false isn't a case of Q => P.
Q.E.D
