Truth Tables for Validity: Using Equations to Determine Satisfaction

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SUMMARY

This discussion focuses on using truth tables to determine the validity of logical equations. The first equation, not(X→(Y→X)), is established as valid due to its inherent truth. The second equation, (X∧(notX→notY))→Y, requires further breakdown for analysis. Participants suggest using truth table identities to simplify complex statements and demonstrate logical equivalences, such as transforming not(A→B) into A∧notB.

PREREQUISITES
  • Understanding of propositional logic
  • Familiarity with truth table construction
  • Knowledge of logical equivalences
  • Experience with logical operators such as AND, OR, and NOT
NEXT STEPS
  • Learn how to construct truth tables for complex logical expressions
  • Study logical equivalences and their applications in propositional logic
  • Explore the use of truth table identities in simplifying logical statements
  • Investigate the implications of validity in logical reasoning
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Students of logic, educators teaching propositional logic, and anyone interested in understanding the foundations of logical reasoning and validity through truth tables.

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I want to use truth tables to show that equations can be satisfied or not, or if they are valid.

not(X→(Y→X))

(X∧(notX→notY))→Y

I would say the first one is valid, because of the not in front of it, it's always true. I don't know about the second one. I don't know how to split them up best to use a truth table. I guess I can/should use:

X, notX, notY, Y, X∧(notX→notY), notX→notY and (X∧(notX→notY))→Y.
 
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Here is one way to look at the problem: If you do a truth table for ## \neg (A \rightarrow B) ##, you will see that it is logically equivalent to the statement ## A \& \neg B ##. So your first statement is equivalent to ## X \& \neg (Y \rightarrow X) ##. But then that is equivalent to ## X \& Y \& \neg X ##. The basic idea is to use truth table identities to transform the complex statement into simpler ones.
 
You can make a truth table with a row for all combinations of y and x. As the columns you use x, y, (y->x), (x->(y->x)) and not(x->(y->x))
Code:
x y    (y->x)    x->(y-x)  not(x->(y-x))
0 0
0 1
1 0
1 1
If there are only ones in al column, the expression is always true, and if there are only zeros, the expression is never true for any x or y.
 

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