TITLE: Equivalence between Peirce's law and Excluded Middle in Intuitionistic logic
QUESTION [5 upvotes]: I'm searching for a intuitionistically valid proof of the formula :


$[((P→Q)→P)→P] ↔ (P \lor \lnot P)$


using the "standard" Hilbert-style axiom system from Kleene [1952], for intuitionistic propositional calculus (with modus ponens).
Following Wiki's proof, I've rewritten the first part :

$\vdash (P \lor \lnot P) → [((P→Q)→P)→P]$

as follows :
1) $P$ --- assumed
2) $((P→Q)→P)→P$ --- from axiom $A → (B → A)$, with $P$ in place of $A$ and $((P→Q)→P)$ as $B$, and 1), by modus ponens
3) $P → [((P→Q)→P)→P]$ --- from 1) and 2) by Deduction Theorem, "discharging" assumption 1)
4) $\lnot P$ --- assumed
5) $(P→Q)$ --- from axiom $\lnot A → (A → B)$ (ex falso quodlibet) and 4), by modus ponens
6) $(P→Q)→P$ --- assumed
7) $P$ --- from 5) and 6), by mp
8) $((P→Q)→P)→P$ --- from 6) and 7), by Deduction Theorem, "discharging" assumption 6)
9) $\lnot P → [((P→Q)→P)→P]$ --- from 4) and 8) by Deduction Theorem, "discharging" assumption 4)
Now that we have derived : $P → [((P→Q)→P)→P]$ (3) and $\lnot P → [((P→Q)→P)→P]$ (9), we use the axiom $(A → C) → ((B → C) → (A ∨ B → C))$ with $P$ in place of $A$, $\lnot P$ in place of $B$ and $((P→Q)→P)→P$ as $C$ to derive, by modus ponens twice :

10) $(P \lor \lnot P) → [((P→Q)→P)→P]$.



I'm not able to manege Wiki's proof of the other part in order to prove the conditional :

$\vdash [((P→Q)→P)→P] → (P \lor \lnot P)$.

In Wiki's proof, as well as in the answer to this post we can find the proof of :

if $\vdash ((P→Q)→P)→P$, then $\vdash P \lor \lnot P$.

The proof use a "substitution instance" of Peirce's law : $(((P \lor \lnot P) \rightarrow \bot) \rightarrow (P \lor \lnot P)) \rightarrow (P \lor \lnot P)$, and this is allowed only if the formula is an axiom or a theorem, which is not in intuitionistic logic. 

For a similar situation, see Jan von Plato, Elements of Logical Reasoning (2013), page 82 :

Given $A \lor \lnot A$, if $\lnot \lnot A$ is assumed, $A$ follows and therefore the implication $\lnot \lnot A \supset A$ follows from $A \lor \lnot A$. An easy proof search in intuitionistic sequent calculus gives a formal derivation for :

$A \lor \lnot A \supset (\lnot \lnot A \supset A)$.

A similar proof search for the converse fails, for there is no derivation of :

$(\lnot \lnot A \supset A) \supset (A \lor \lnot A)$.

[...]
The overall situation is that if $\vdash \lnot \lnot A \supset A$, then $\vdash A \lor \lnot A$, but not $\vdash (\lnot \lnot A \supset A) \supset (A \lor \lnot A)$.


Thus, the Question is : is it possible that 


if $\vdash ((P→Q)→P)→P$, then $\vdash P \lor \lnot P$, 
but $\nvdash [((P→Q)→P)→P] → (P \lor \lnot P)$

REPLY [4 votes]: The equivalence between Peirce's formula and the excluded middle is sometimes claimed but because
$$(((A \to B) \to A) \to A) \to (\lnot A \lor A)$$
is not an intuitionistic theorem, this equivalence fails in intuitionistic logic.
But  this instance of Peirce's formula:
$$((A \to \bot) \to A) \to A$$ i.e.
$$(\lnot A \to A) \to A$$
is intuitionistically equivalent to the classical double negation elimination; i.e.
$$\lnot \lnot A \to A$$