I understand the general concept, that if you define a statement X such that X = X $\Rightarrow$ Y, you can prove that X is true regardless of Y so you can then prove any statement. What I don't quite understand is how one proves that X is true.
I have tried to prove it using the three main ways of proving an implication, contradiction, contrapositive, and direct, but the last two methods are unsatisfactory.
Contradiction:
Suppose X is true and Y is false.
Then X $\Rightarrow$ Y is false.
But X = X $\Rightarrow$ Y.
X cannot be both true and false. So X cannot be true when Y is false.
So X $\Rightarrow$ Y is true, so X is true.
QED
This way I understand and seems fine, what trips me up is direct and contrapositive, with direct assuming what you are trying to show and contrapositive showing that X is both true and false.
Direct:
Suppose X is true.
Then since X = X $\Rightarrow$ Y is true, and X is true, Y is true.
So X $\Rightarrow$ Y is true, and hence X is true.
QED
This seems to assume what we are trying to prove.
Contrapositive:
Suppose Y is false.
If X is true, then X $\Rightarrow$ Y is false.
But X = X $\Rightarrow$ Y so X is false.
X cannot be both true and false, so X is false.
Therefore X $\Rightarrow$ Y is true, so X is true.
QED
This is troubling because X has been shown to be both true and false.
I guess what I am asking is what is the accepted way to prove that a statement X = X $\Rightarrow$ Y is true to construct Curry's paradox, and why do some proof methods seem incorrect or contradictory?
Best Answer
First of all, $X \leftrightarrow (X \rightarrow Y)$ implies both $X$ and $Y$, and the whole point of the Curry Paradox is to show that $Y$ is true ... $X$ is just a 'helper'.
Second, you can show that $X \leftrightarrow (X \rightarrow Y)$ implies both $X$ and $Y$ in various ways:
First, truth-table:
\begin{array}{cc|c} X&Y&X \leftrightarrow (X \rightarrow Y)\\ \hline T&T&T\\ T&F&F\\ F&T&F\\ F&F&F\\ \end{array}
We see that $X \leftrightarrow (X \rightarrow Y)$ is only true in row 1 ... when $X$ and $Y$ are both true
Second, algebra:
$X \leftrightarrow (X \rightarrow Y) \overset{Equivalence}\Leftrightarrow$
$(X \rightarrow (X \rightarrow Y)) \land ((X \rightarrow Y) \rightarrow X) \overset{Implication}\Leftrightarrow$
$(\neg X \lor (\neg X \lor Y)) \land (\neg (\neg X \lor Y) \lor X) \overset{Association, DeMorgan}\Leftrightarrow$
$(\neg X \lor \neg X \lor Y) \land ((X \land \neg Y) \lor X) \overset{Idempotence, Absorption}\Leftrightarrow$
$(\neg X \lor Y) \land X \overset{Reduction}\Leftrightarrow$
$Y \land X$
Third, formal proof: