Showing that the following isn't complete

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SUMMARY

The discussion centers on proving the incompleteness of the logical operators \left\{ \neg,\equiv\right\} in expressing the implication operator (->). The user suggests demonstrating that any proposition formed with the operators ~ (negation) and <-> (biconditional) must exhibit an even property in its truth table, which can be established through mathematical induction. Since the implication operator exhibits an odd property, it cannot be represented using only ~ and <->. The user confirms they successfully completed this proof and expresses gratitude for the assistance received.

PREREQUISITES
  • Understanding of propositional logic and operators such as negation (~) and biconditional (<->).
  • Familiarity with truth tables and their properties.
  • Knowledge of mathematical induction as a proof technique.
  • Basic concepts of logical implication (->) and its properties.
NEXT STEPS
  • Study the properties of logical operators in propositional logic.
  • Learn how to construct and analyze truth tables for various logical expressions.
  • Explore mathematical induction and its applications in formal proofs.
  • Investigate other logical operators and their expressiveness in propositional logic.
USEFUL FOR

Students of mathematics, logicians, and anyone interested in the foundations of propositional logic and proof techniques.

Palindrom
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Let's have a look at [tex]\left\{ \neg,\equiv\right\}[/tex]. How could one show that this isn't complete?

I've tried finding some sort of invariance that propositions built with these might have, but I couldn't find anything... I'm going crazy! :smile:
 
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The set {~,<->} is inadequate.

Proof? You might try to prove that operator -> can't be expressed by any combination of the operators ~ and <->.

A start would be to show that the truth table for any proposition that is made up of 2 or more propositional
symbols and only the operators ~ and <-> must have an even number of ones and an even number of zeros in its last column (call it the even property). I think it's pretty clear this would have to be done by induction.

Then argue that since -> has the odd property it can't be expressed using only ~ and <->.

(The operator -> could be replaced by either V or &.)
 
Last edited:
Thanks, I did do this eventually, and just came back here now to thank anyone that might have answered.

So thanks!
 

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