For formal logic proposition A, if true-valued function of A is T(A)=0, the false, then T(T(A))=? If T(T(A))=0 then implies T(A)=1, the true; if T(T(A))=1, then implies T(A)=0, Is this a real paradox?
T(T(A))=0 is equal to T(A)=1 or 0 is not correct together at same time.
If we see T(A) as a NEW proposition, then we can apply chained true-valued function to obtain the T(T(A)) .
The key problems are the two cases:
1) T(A)=0, and T(T(A))=0;
2) T(A)=1, and T(T(A))=0.
This is Liar paradox! PARADOX! It must leads to a contradiction! T(T(A)) is a judgement for T(A), i.e. T(A) is true or false. Chained true-valued is an operator that well-known seen. While we take a judgment to T(A) with true or false, then we obtained T(T(A)). If T(T(A))=0, we say proposition A is falsehood, and if T(T(A))=1 then we say proposition A is truth, but it is impossible because proposition A is from a liar. New operator has not been produced yet, all troubles have not been born yet!
OK?
See my new reply, please.I am glad to your arguments! Thanks! I hope you state continuously your opinions.
If John, as a liar, gives a proposition A, and judges its truth by true-valued function T(A), then Mary re-judges the NEW proposition T(A) by T(T(A)), all of these have been ruled in Boolean Algebra as well as they did, What sort ill-formed is made out? Isn't T(A) a valid proposition? Isn't T(T(A)) a valid true-valued function? Mary is not a horse, however a horse is not a girl. In fact, chained true-valued function is a well-known operator in formal logic, and we can have, if necessary, not only T(T(A)), but also T(T(T(A))), T(T(T(T(A)))), and so on, isn't it?
Thanks for your aguments!
Can you argue with me to the two other paradoxes written by me?
T is an operator of true-valued operation, A is the proposition.Suppose T(A) is expressed by sentence as: proposition A is false, such a sentence isn't a proposition? Why do we see T(A) as Not a proposition, although it is formed mathematically?
My domain is also X={0,1}, T(T(A)) indeed is a mapping, or an operator, makes f:A->X={0,1}, and f:T(A)->X={0,1}, THIS A PARADOX! It must lead to contradiction.
"Mary is a horse" the true-valued function of this proposition is 0.
If proposition A="Mary is a horse", then T(A)=0. Why do we can say T(A)=0 is not belong to {0,1}? The T is an operator, or a mapping, this is obviously!
T is a true-valued oprator operates on the proposition A, therefor we can obtain the T(A), the true-valued function . Although T(A) is belong to {0,1}, but T(A) is also a proposition, thus it can be operated on second operation by the operator, and third operation by the operator, and so on. This is the key comment of your argment view. This chained true-valued function will be fitted by whom unknown to it.
"Mary is horse" isn't a proposition? And suppose A="Mary is horse", isn't a proposition? Then," "Mary is horse" is false" isn't a proposition? T(A)=0, or 1, why does it can be seen as not a proposition! Chained true-valued function, such a simple concept actually consumes our so much understanding! This really strikes us! I am glad to your consistence! Go on!
Proposition by sentence, or by form mathematical, their mode treated by has no an essential distinction.
I hope discussion for other two paradoxes!
An operator of chained true-valued function not only produce a paradox, but also produce confusions.
If proposition is from a liar, then the true-valued function should be neither 0 nor 1, might be the third value, of that is reason for multiple true-valued function.
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