Problem understanding why $P \implies (Q \implies P) \equiv T$ I've been through the truth table and I can see how it works but I can't exactly understand why. The proof presented in the book (Logical Reasoning: A First Course by Nederpelt and Kamareddine) says that the derivation is as follows:
$$1. \{Assume: P\}$$
$$2. \{Assume: Q\}$$
$$3. \{Valid(1): P\}$$
$$4. \{By(2) and (3): Q \implies P\}$$
$$5. \{By(1) and (4): P \implies (Q \implies P)\}$$
 A: It is a tautology, i.e. a formula identically true (i.e. a formula $\mathcal A$ such that $\mathcal A \equiv T$), as you as verified using truth table method.
Thus, by Completeness Theorem for propositional logic it must be provable.
In many proof system in Hilbert-style (for propositional logic) it is an axiom.
We can prove it with Natural Deduction.

Proof :
i) $P$ - assumed
ii) $Q$ --- assumed
iii) $P,Q \vdash P$ --- from i) and ii)
iv) $P \vdash (Q \rightarrow P)$ --- from iii) by $\rightarrow$-I

v) $\vdash P \rightarrow (Q \rightarrow P)$ --- from iv) by $\rightarrow$-I.


For an explanation, see Jan von Plato, Elements of Logical Reasoning (2013), page 22 :


There is a limiting case of a derivation in which an assumption $A$ is made. It is at the same time a derivation of the conclusion $A$ from the assumption $A$, as in:
    
    
*
    
*$A$ : hypothesis
    
*$A \rightarrow A$ : 1,$\rightarrow$-I
In terms of the derivability relation, the hypothesis on line 1 can be written as $A \vdash A$ and line 2 as $\vdash A \rightarrow A$.
Consider as another case $\vdash A \rightarrow (B \rightarrow A)$. Verbally, if we assume $A$, then $A$ follows under any other assumption $B$ :
    
    
*
    
*$A$ : hypothesis
    
*$B \rightarrow A$ : 1,$\rightarrow$-I
    
*$A \rightarrow (B \rightarrow A)$ : 1–2,$\rightarrow$-I
This does not look particularly nice: We have closed an assumption $B$ that was not made. But if we say that an assumption was used $0$ times, the thing starts looking more reasonable. [...] we can say that assumption $B$ in the derivation of $A \rightarrow (B \rightarrow A)$ was used vacuously.


A: Prove $P\implies [Q\implies P]$, or equivalently $\neg[P\land [Q\land \neg P]]$.
Suppose to the contrary that $P\land [Q\land \neg P]$ is true and obtain the obvious contradiction $P \land \neg P$.

Everything, even that which is false, implies that which is true. 


Similarly, we could also prove that $P\implies [\neg P \implies Q]$, or equivalently $\neg[P \land [\neg P \land \neg Q]]$.

Everything, even that which is false, follows from that which is false.

A: This is a proof by Natural Deduction. The variables $Q$ and $P$ have been discharged respectively in steps $4$ and $5$. But the semantics are thus: 
This essentially means If $P$ then $[$ If $Q$ then $P$ $]$. Now suppose $P$ is true. Then the statement If $Q$ then $P$ is true regardless of what the statement $Q$ is. Because $P$ if true if $Q$ is true because $P$ is true anyway. Now we have just concluded that if $P$ is true then the statement $[$ If $Q$ then  $P$ $]$ is true. Whence, we can conclude that If $P$ then $[$ If $Q$ then $P$ $]$ is a true statement. 
