Ytdyklly

I am stuck at showing how if:

$P rightarrow Q$ then this implies ($Q rightarrow R) rightarrow (P rightarrow R)$

I know that if we assume $P$ is true then $Q$ also must be true.
Therefore if $Q$ is true then $R$ must be true. And because $Q$ is true because $P$ is true so therefore $R$ must also be true.

However it would be appreciated if someone could help me understand this why is the case and what proof technique it is.

This property is called transitivity of implication.

You pretty much gave a proof in your question, but here's a proof written out in slightly more precise terms.

Suppose $P Rightarrow Q$ is true. To prove $(Q Rightarrow R) Rightarrow (P Rightarrow R)$, you need to assume $Q Rightarrow R$ and derive $P Rightarrow R$. So assume that $Q Rightarrow R$ is true. To prove $P Rightarrow R$ is true, you need to assume $P$ is true and derive $R$. So assume $P$ is true. All we have to do now is prove that $R$ is true.

At this point, we're assuming that $P Rightarrow Q$, $Q Rightarrow R$ and $P$ are all true. So:

• Since $P$ and $P Rightarrow Q$ are true, we have that $Q$ is true; and


• Since $Q$ and $Q Rightarrow R$ are true, we have that $R$ is true.

So we're done.

You can also prove this somewhat verbosely with a truth table... (I have named my intermediate steps just because the table was too wide to display nicely.)

$$
begin{array}{|c c c|c|c|c|c|c|}
P & Q & R & P implies Q & Q implies R & P implies R & & \
& & &A&B&C&B implies C & A implies (B implies C) \
\
T & T & T & T & T & T & T & T\
T & T & F & T & F & F & T & T\
T & F & T & F & T & T & T & T\
T & F & F & F & T & F & F & T\
F & T & T & T & T & T & T & T\
F & T & F & T & F & T & T & T\
F & F & T & T & T & T & T & T\
F & F & F & T & T & T & T & T\
end{array}
$$

It’s a sort of transitivity. $Pimplies Q$ is true. So if $Qimplies R$, then $Pimplies Qimplies R$. I don’t think there is a name for this. It’s just multiple implications that give a result.

Proving this is simple. Given $Qimplies R$, we want to demonstrate $Pimplies R$.
So let’s suppose that $P$ is true. So, $Q$ is true by $(Pimplies Q)$ (it’s called the “Modus ponens”). But $Q$ is true, so $R$ is true.
QED.

If you want a conceptual proof, you just need to know how to prove an implication. How do you prove $Arightarrow B$? Assume $A$ is true, then show that $B$ follows (hoping you have some other information laying around to do this using $A$.

• Step 1 To show:
  $(P rightarrow Q) rightarrow ((Q rightarrow R) rightarrow (P rightarrow R))$,
  assume $(P rightarrow Q)$ and show $(Q rightarrow R) rightarrow (P rightarrow R)$.




• step 2 To show $(Q rightarrow R) rightarrow (P rightarrow R)$, assume $Q rightarrow R$ and then show $P rightarrow R$.




• step 3 To show $P rightarrow R$, assume $P$ and then show $R$.

At each stage we're assuming something so by this last step we've assumed a whole bunch that can help us to show $R$.

Using $P$, assumed true from the last step and $Prightarrow Q$ assumed true from step 1, $Q$ follows, via modus ponens.

Using this $Q$ and $Qrightarrow R$ assumed true from step 2, $R$ follows, again via modus ponens.

And we're done. In short, a chain of implications unraveled from left to right.

$ARightarrow B$ can be written as $neg Avee B.$ Therefore,
$$
(QRightarrow R)Rightarrow(PRightarrow R) \
=neg(neg Qvee R)vee (neg Pvee R) \
= (Q wedge neg R)vee (neg Pvee R) \
= (neg P vee Qvee R)wedge (neg P vee R vee neg R) \
= (neg P vee Qvee R) \
= (neg P vee Q)vee R \
= (PRightarrow Q) vee R
$$