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[edit] Correctness of Duplication

Forgive me if I'm wrong, but shouldn't Axiom 1 be:

$\bullet\; \forall x \lbrace [\varphi(x) \rightarrow \psi(x)] \land P(\varphi){\color{Red}\rbrace} \rightarrow P(\Psi)\rbrace$

It seems that we're closing a non-opened square bracket. Since this is a quite famous result, and since I'm not entirely familiar with logical notation (though I don't see this happen anywhere else in the proof), I'm apprehensive about changing this on my own. Error792 07:32, 10 January 2007 (UTC)

Shouldn't it be

P (ψ)

(i.e., lowercase psi)? Isn't this axiom saying: if phi(x) implies psi(x), and phi is positive, then psi is positive? --Taejo|대조 20:46, 13 May 2007 (UTC)

You're absolutely right, I capitalized the psi by accident. Error792 03:26, 23 June 2007 (UTC)

[edit] Proof

In wiki notation:

Ax. 1. $\bullet\; \forall x \lbrace [\varphi(x) \rightarrow \psi(x)] \land P(\varphi)] \rightarrow P(\Psi)\rbrace$

Ax. 2. $P(\neg \varphi) \leftrightarrow \neg P(\varphi)$

Th. 1. $P(\varphi) \rightarrow \Diamond\; \exists x\; [\varphi(x)]$

Df. 1. $G(x) \leftrightarrow \forall \varphi[P(\varphi) \rightarrow \varphi(x)]$

Ax. 3. $P (G)$

Th. 2. $\Diamond\; \exists x\; G(x)$

Df. 2. $\varphi\;\mathrm{ess}\;x \leftrightarrow \varphi(x) \land \forall\psi\lbrace\psi(x) \rightarrow \bullet\; \forall x[\varphi(x) \rightarrow \psi(x)]\rbrace$