Learning the rules for Natural Deduction. I proved the following statement with the following proof :
( P ∨ Q ) ∧ R ⊢ (P ∧ R) ∨ (Q ∧ R)
( P ∨ Q ) ∧ R ⊢ (P ∧ R) ∨ (Q ∧ R)
1. ( P ∨ Q ) ∧ R - Assumption 0
2. ( P ∨ Q ) - ∧ - Elimination - 1
3. R - ∧ - Elimination - 1
4. P - ∨ - Elimination - 2
5. Q - ∨ - Elimination - 2
6. P ∧ R - ∧ - Introduction - 3,4
7. Q ∧ R - ∧ - Introduction - 5,3
8. (P ∧ R) ∨ (Q ∧ R) - ∨ - Introduction - 6,7
However, looking at the model answer the proof is given as
(b) One direction of a distribution law:
$$[(p \lor q) \land r \vdash(p \land r) \lor(q \land r)]$$
\begin{array}{l|lr}\hline
1 & (p \vee q) \land r & \text { assumption, 0 } \\
2 & p \lor q & \land-\mathrm{E}, 1 \\
3 & r & \land-\mathrm{E}, 1 \\
4 & p \vdash(p \land r) \lor(q \land r) & \\
4.1 & p & \text { assumption, 4 } \\
4.2 & p \land r & \land-\mathrm{I}, 4.1,3 \\
4.3 & (p \land r) \lor(q \land r) & \lor-\mathrm{I}, 4.2 \\
5 & p \implies((p \land r) \lor(q \land r)) & \implies-1,4 \\
6 & q \vdash(p \wedge r) \vee(q \wedge r) & \\
6.1 & q & \text { assumption, 6 } \\
6.2 & q \land r & \land-\mathrm{I}, 6.1,3 \\
6.3 & (p \land r) \lor(q \land r) & \lor-\mathrm{I}, 6.2 \\
7 & q \implies((p \land r) \lor(q \land r)) & \implies-1,6 \\
8 & (p \land r) \lor(q \land r) & \lor-\mathrm{E}, 2,5,7
\end{array}
Are both proofs valid, or is mine incorrect? The second proof appears far more elaborate
Best Answer
Your proof is definitely not correct. To be precise: lines 4 and 5 are incorrect.
You basically try to infer $P$ from $P \lor Q$, that that does not follow. If I tell you that my shirt is either red or blue, can you infer that it is therefore red? No.