Example from "Beginner Logic" (E.J.Lemmon)

Question

$\newcommand{\ass}[2]{ #1 \qquad &(#1)#2 \qquad & A }$ $\newcommand{\di}[4]{ #4 \qquad &(#1)#2 \lor #3 \qquad & #4 \lor I }$ $\newcommand{\mpp}[4]{ #4 \qquad &(#1)#2 \qquad & #3MPP }$ $\newcommand{\de}[5]{ #5 \qquad &(#1) #2 \lor #3 \qquad & #4 \lor E }$ $\newcommand{\cp}[5]{ #5 \qquad &(#1) #2 \to #3 \qquad & #4 CP }$

Hello, I'm currently learn logic as a complete beginner. This example from the book for explaining Disjunction Elimination has the following proof:

$$Q \to R \vdash (P \lor Q) \to (P \lor R) \\ \begin{align} \ass{1}{Q \to R} \\ \ass{2}{P \lor Q} \\ \ass{3}{P} \\ \di{4}{P}{R}{3} \\ \ass{5}{Q} \\ \mpp{6}{R}{1,5}{1,5} \\ \di{7}{P}{R}{6} \\ \de{8}{P}{R}{2,3,4,5,7}{1,2} \\ \cp{9}{(P \lor Q)}{(P \lor R)}{2,8}{1} \end{align} $$

As you can see "$(6)R$" is derived from 1,5MPP, but could it be derived from an assumption from Disjunction Introduction in $(4) P \lor R$ ? My guess is that is better to use MPP since, as the author states, in case of Disjunction Introduction "Clearly the conclusion will in general be weaker than the premises" and such would be deriving an assumption from such conclusion. Thanks

Answer 1

$\newcommand{\ass}[2]{ #1 \qquad &(#1)#2 \qquad & A }$ $\newcommand{\di}[4]{ #4 \qquad &(#1)#2 \lor #3 \qquad & #4 \lor I }$ $\newcommand{\mpp}[4]{ #4 \qquad &(#1)#2 \qquad & #3MPP }$ $\newcommand{\de}[5]{ #5 \qquad &(#1) #2 \lor #3 \qquad & #4 \lor E }$ $\newcommand{\cp}[5]{ #5 \qquad &(#1) #2 \to #3 \qquad & #4 CP }$

No, because disjunction introduction means that if either $P$ or $R$ (or both) then we can conclude that $P \lor R$. So, if $P \lor R$ and $R$ we could still have $\lnot P$.

So, the inverse is true: $P \vdash P \lor R$ but not necessarily $P \lor R \vdash P$. We would get there in this case: $$P \lor R, \ \lnot R \vdash P.$$