TITLE: Can Peirce's Law be proven without contradiction?
QUESTION [6 upvotes]: Good evening, I heard the proof by contradiction is required for Peirce's law. AFAIK, truth tables are not related directly to proofs by contradiction, and if of an operation $\text {op}$ we have a truth table 
P(p) P(q)   P(p op q)
  0    0        1
  0    1        1
  1    0        1
  1    1        1

such that $P$ is the predicate, or logical/truth value of a proposition, then
$$
\forall p\forall q, P(p\ \text{op}\ q)=1\Leftrightarrow\forall p\forall q, p\ \text{op}\ q
$$
So isn't 
(p    q   if p, then q  if p, q; and then p    modus ponens           Peirce's Law)
P(p) P(q)    P(p → q)     P((p → q) → p)   P(((p → q) → p) → q)   P(((p → q) → p) → p)
  0    0         1              0                   1                      1       
  0    1         1              0                   1                      1       
  1    0         0              1                   1                      1       
  1    1         1              1                   1                      1       

a proof of Peirce's law?
(you can make the maths yourself, noting this is classical logic and that $p \to q \Leftrightarrow \lnot p\lor q$; and all works without admiting any proof by contradiction at all)

REPLY [2 votes]: I think that the statement "the proof by contradiction is required for Peirce's law" must be interpreted in a different way.
I refer to these previous post about Implicational calculus.
My starting point is the answer by @Doug Spoonwood : according to Lukasiewicz' axiomatisation, one axiom is enough to derive all the "implicational" tautologies.
The issue is that in this way there is place for negation ($\lnot$) in the system.
The easiest way is to introduce the falsum ($\bot$); with it we may use $\lnot A$ as an abbreviation of $A \rightarrow \bot$.
In the previous post, I argued that, in order to derive the classical laws of negation, like Double Negation, we need Peirce's law plus the Ex Falso Quodlibet axiom :

$\vdash \bot \rightarrow A$.

Also in order to prove the equivalence between Peirce's law and the Law of Excluded Middle, we need the symbol $\bot$ as primitive and the EFQ axiom.