Logical equivalencies involving ifs and nors

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SUMMARY

The discussion focuses on finding a compound proposition logically equivalent to p → q using only the logical operator NOR (↓). The solution involves transforming the implication into a series of logical equivalencies, ultimately deriving ¬((p ↓ p) ↓ q) through the application of the NOT function derived from p NOR p. The key steps include recognizing that ¬p can be expressed as p ↓ p, allowing for the manipulation of the original proposition into a form that utilizes only the NOR operator.

PREREQUISITES
  • Understanding of logical operators, specifically NOR (↓).
  • Familiarity with logical equivalencies and implications in propositional logic.
  • Knowledge of De Morgan's laws as they apply to logical expressions.
  • Ability to manipulate compound propositions and apply transformations systematically.
NEXT STEPS
  • Study the properties and applications of the NOR operator in propositional logic.
  • Learn about De Morgan's laws and their implications in logical equivalencies.
  • Explore additional logical equivalencies involving implications and conjunctions.
  • Practice converting various logical expressions into forms using only NOR and other operators.
USEFUL FOR

Students of logic, mathematicians, and anyone interested in deepening their understanding of propositional logic and logical equivalencies.

nicnicman
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Can anyone solve this step by step, so I can see how it's done? I've been at for a while now and can't seem to get it. Here's the problem:

Find a compound proposition logically equivalent to p → q using only the logical operator ↓.

Thanks for the help. I'm really trying to get this stuff, but it's not coming easy.
 
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The trick is to obtain a NOT function by p NOR p. That breaks the symmetry.
 
Thanks for the suggestion, I finally got it.

If anyone's interested:

p → q
≡ ¬p ∨ p this is one of the common equivalencies given in my book by Deitel
≡ ¬(¬p ↓ p) this equivalency was found in a previous exercise
≡ ¬((p ↓ p) ↓ q) by ¬p ≡ p ↓p
≡ ((p ↓ p) ↓ q) ↓ ((p ↓ p) ↓ q) by ¬p ≡ p ↓p
 

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