How to prove (p → q) ∨ (q → r) ≡ (p ∧ q) → r

Question

My attempt at this mathematical demonstration was as follows:

(p → q) ∨ (q → r) ≡ (~p ∧ q) ∨ (q → r) ≡ q ∨ (p → r) ≡ (q ∨ p) → r ≡ (p ∨ q) → r

Answer 1

Your second step seems off, you went from $p\rightarrow r$ to $\lnot p\land q$, it is not clear how q appeared from r. Instead, try this, using the fact that $a\rightarrow\ b$ means $\lnot a \lor b$:

$$\begin{split}(p\rightarrow r)\lor (q\rightarrow r)\\ (\lnot p \lor r)\lor(\lnot q \lor r)\\ (\lnot p \lor \lnot q)\lor (r\lor r)\\ (\lnot p \lor \lnot q)\lor r\\ \lnot(p\land q)\lor r\\ (p\land q)\rightarrow r\end{split}$$

Answer 2

Here's one way to prove it that's nonstandard, it uses $[ a \, b \, c \,\cdots]$ as a form of nand that takes an arbitrary number of arguments. This is a linear notation for an existential graph and leverages the fact that nand is a sole sufficient operator for classical logic.

$$ [ a \, b \, c \, \cdots ] \;\;\text{is equivalent to}\;\; \lnot(a \land b \land c \land \cdots) $$

Repetition of the arguments and order of the arguments to a single $[\cdots]$ does not affect the truth value of the expression.

Inserting or removing a double negation also does not affect the truth value of an expression.

For example,

$$ [a \, b] \;\; \text{is equivalent to} \;\; [a \, [[b]]] $$

Note that

$$ p \to q \;\;\text{is equivalent to}\;\; [p \, [q]] $$

$$ p \lor q \;\;\text{is equivalent to}\;\; [[p]\,[q]] $$

And the definition of and, by way of double negation elimination

$$ p \land q \;\;\text{is equivalent to}\;\; [[p \, q]] $$

The first expression is equivalent to

$$ [[[p \, [r]]] \, [[q \, [r]]]] $$

We begin by removing the double negation

$$ [ p \, [r] \, q \, [r] ] $$

Remove the repeated argument

$$ [p \, q \, [r]] $$

Insert a double negation

$$ [[[p \, q]] \, [r]] $$

Apply the definition of $\land$ and the definition of $\to$

$$ (p \land q) \to r $$

Answer 3

These statements are not equivalent to one another. In fact, the LHS expression is a tautology! To see this, note that if $q$ is true, then $p \to q$ is true (anything implies a true statement) and if $q$ is false, then $q \to r$ is true (a false statement implies anything).

However, the RHS is not a tautology. Specifically, it’s false if $p$ and $q$ are true and $r$ is false.

Hope this helps!