Question

((𝑝 → 𝑞) ∧ (𝑞 → 𝑟)) → (𝑝 → 𝑟)

Answer 1

Let us construct the truth table of the following proposition.

$f=((𝑝 → 𝑞) ∧ (𝑞 → 𝑟)) → (𝑝 → 𝑟)$

$\ \begin{array}{||c|c|c||c|c|c|c|c||} \hline\hline p & q & r &𝑝 → 𝑞 & 𝑞 → 𝑟 & (𝑝 → 𝑞) ∧ (𝑞 → 𝑟) & 𝑝 → 𝑟 & f\\ \hline\hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1\\ \hline 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1\\ \hline 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline\hline \end{array}$

It follows that the formula $((𝑝 → 𝑞) ∧ (𝑞 → 𝑟)) → (𝑝 → 𝑟)$ is a tautology.