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

by VinnyCee
Tags: conditional, statement, tables, tautology, truth
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VinnyCee
#1
Sep10-07, 05:37 PM
P: 492
1. The problem statement, all variables and given/known data

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



2. Relevant equations

DeMorgan's Laws, etc.



3. 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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JonF
#2
Sep10-07, 08:09 PM
P: 617
there is an error in your first line

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


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