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

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VinnyCee
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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