Show that the conditional statement is a Tautology without using truth tables

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SUMMARY

The conditional statement \(\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q\) is proven to be a tautology without using truth tables. The proof utilizes DeMorgan's Laws and logical equivalences, transforming the expression into \(\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\vee\,q\) and further simplifying it to demonstrate that it holds true under all interpretations. The discussion highlights the importance of recognizing logical equivalences, specifically that \(a \rightarrow b\) is logically equivalent to \(\neg a \vee b\).

PREREQUISITES
  • Understanding of propositional logic
  • Familiarity with DeMorgan's Laws
  • Knowledge of logical equivalences
  • Basic skills in symbolic manipulation
NEXT STEPS
  • Study logical equivalences in depth
  • Explore advanced applications of DeMorgan's Laws
  • Practice symbolic logic proofs without truth tables
  • Learn about other forms of tautologies in propositional logic
USEFUL FOR

Students of mathematics, particularly those studying logic, computer science students focusing on algorithms, and educators teaching propositional logic concepts.

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Homework Statement



Show that [tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q[/tex] is a tautology without using truth tables.



Homework Equations



DeMorgan's Laws, etc.



The Attempt at a Solution



[tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\longrightarrow\,q[/tex]

by. EX 3 (see EX 8)

[tex]\left[\neg\,p\,\wedge\,\left(p\,\vee\,q\right)\right]\,\vee\,q[/tex]

[tex]\left[p\,\wedge\,\neg\,\left(p\,\vee\,q\right)\right]\,\vee\,q[/tex]

[tex]\left[p\,\wedge\,\left(\neg\,p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

[tex]\left[\left(p\,\wedge\,\neg\,p\right)\,\wedge\,\left(p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

[tex]\left[F\,\wedge\,\left(p\,\wedge\,\neg\,q\right)\right]\,\vee\,q[/tex]

Now what?
 
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there is an error in your first line

a -> b is logicaly equivlent to ~a or b
 

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