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

Click For Summary
SUMMARY

The logical equivalence of the expression (P -> Q) or (P -> R) to P -> (Q or R) is confirmed to be incorrect based on the analysis presented in the forum discussion. The user attempted to derive the equivalence through various logical transformations but found inconsistencies, particularly in the transition from (P -> Q) or (P -> R) to (P or ¬Q) or (P or ¬R). The correct interpretation involves recognizing that the expressions do not yield the same results under logical operations.

PREREQUISITES
  • Understanding of propositional logic and logical operators
  • Familiarity with logical equivalences and transformations
  • Knowledge of truth tables and their application in logical proofs
  • Experience with symbolic notation in logic
NEXT STEPS
  • Study the principles of logical equivalence in propositional logic
  • Learn how to construct and analyze truth tables for complex expressions
  • Explore the use of logical identities and their proofs
  • Investigate the implications of De Morgan's laws in logical expressions
USEFUL FOR

Students of mathematics, logic enthusiasts, and educators looking to deepen their understanding of logical equivalences and their applications in formal proofs.

The Subject
Messages
32
Reaction score
0
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
(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)
and
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)

From what I've done its seems like they're not equivalent ?
 
Physics news on Phys.org
The Subject said:
I tried
(P \to Q) \lor (P \to R) \\<br /> (P \lor \neg Q) \lor (P \lor \neg R)
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!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K