Hello all.
I know the last expresion is not allways true(by making examples), but it can be proved that is true!
We know that to being the a true expression, both the expressions right to the (first) equalency sign must be true. So we have:
It can be proved by a similar way that:
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What do examples contradict to this theorem?
Does the proof have a problem?
Thanks.
By propositional calculus you indeed have that but it is NOT true that .
So where you have (in your original expression): the equivalence does not hold because you do not have .
That's where you have gone wrong.
No, you are wrong!!
Your statement ( is not true) holds whenever we don't know about validity of Q.(i mean we don't know it is true or false).
But i've said clearly that here (wich is a true statement, and here it is ).
So the equvalency sign is i have chosen is TRUE.
Because and .
First of all we cannot mix a syntactical with semantical proof ,a thing that you have done in your proof.
In a syntactical proof one uses the rules of logic without using true or false values for parts of formulas.
In a semantical proof one uses the true tables .
Mixing the two the results are wrong conclusions.
Besides that, to have: we must have that :
logicaly implies
................................................OR.......................
The conditional : is a tautology.
But it is not .
Hence : does not hold .
Thus your proof is wrong
AND another more obvious thing: In the backwards proof ( the one starting with xεΑ ) When you substitute T = xεΒ ,this is wrong because you do not know now the value of xεΒ
I think I understand now (of course, "T" is "True" = "tautology")
IF it is the case that , which it only is if is ALWAYS an element of , then the equivalence indeed holds. But then that means which is indeed equivalent to .
But first you have to take as a truth that ALL THE TIME, WHATEVER and may be.
WHICH IS RUBBISH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!