Proofs aren't always easy to write out but I'll attempt a lengthier proof that I've recently seen. The assertion is that the proposition [p
→ (q → r)] → [(p → q) → (p → r)] is a tautology. We can approach this in a couple of different ways. One way would be to create a truth table, which I'll demonstrate below and the other will be to do it with inference laws. I'm not going to list out all of the inference laws, but will instead just work out the example.

The truth table should display True for all of the different combinations and as can be seen below it does.

p q r q → r p → (q → r) p → q p → r (p → q) → (p → r) [p → (q → r)] →

[(p → q) → (p → r)] T T T T T T T T T T T F F F T F

The truth table should display True for all of the different combinations and as can be seen below it does.

p q r q → r p → (q → r) p → q p → r (p → q) → (p → r) [p → (q → r)] →

[(p → q) → (p → r)] T T T T T T T T T T T F F F T F