Prove Logical Equivalence of P->(Q or R)

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The Subject
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From the text it says (P -> Q) or (P -> R) is equivalent to P -> (Q or R)

I tried to see if this is true so I tried
[tex](P \to Q) \lor (P \to R) \\<br /> (P \lor \neg Q) \lor (P \lor \neg R) \\<br /> P \lor \neg Q \lor \neg R \\<br /> P \lor \neg(Q \land R) \\<br /> P \to (Q \land R)[/tex]
and
[tex]P \to (Q \lor R) \\<br /> P \lor \neg(Q \lor R ) \\<br /> P \lor (\neg Q \land \neg R) \\<br /> (P \lor \neg Q) \land (P \lor \neg R) \\<br /> (P \to Q) \land (P \to R)[/tex]

From what I've done its seems like they're not equivalent ?
 
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The Subject said:
I tried
[tex](P \to Q) \lor (P \to R) \\<br /> (P \lor \neg Q) \lor (P \lor \neg R)[/tex]
The second line does not follow from the first.

I think what you meant to write for the second line was
$$(\neg P\vee Q)\vee (\neg P\vee R)$$
which is not the same thing.
 
andrewkirk said:
The second line does not follow from the first.

I think what you meant to write for the second line was
$$(\neg P\vee Q)\vee (\neg P\vee R)$$
which is not the same thing.
AHHHH thank you!