Help with prenex normal form
LEQV = Logically equivalent
I need help changing
∀xA(x) ^ ∀xB(x) ^ ∀xC(x) → ∀xD(x)
into PNF (such that the only connectives in the quantifier free portion have to be →)
∀xA(x) ^ ∀xB(x) ^ ∀xC(x) → ∀xD(x)
LEQV
¬ ( ∀xA(x) ^ ∀xB(x) ^ ∀xC(x) ) v ∀xD(x) [→ Law]
LEQV
( ∃x¬A(x) v ∃x¬B(x) v ∃x¬C(x) ) v ∀xD(x) [Duality of quantifiers law]
LEQV
∃x1∃x2∃x3∀x ( ( ¬A(x1) v ¬B(x2) v ¬C(x3) ) v D(x) ) [pulling quantifiers out]
LEQV
∃x1∃x2∃x3∀x ( ¬ ( A(x1) ^ B(x2) ^ C(x3) ) v D(x) ) [pulling¬ out]
LEQV
∃x1∃x2∃x3∀x ( ( A(x1) ^ B(x2) ^ C(x3) ) → D(x) ) [reverse → Law ]
I got here but I need the only connectives in the quantifier free portion to be →, right now the last step includes ^ symbols...
Thanks
Re: Help with prenex normal form
Re: Help with prenex normal form
Hmm is this right?
∀xA(x) ^ ∀xB(x) ^ ∀xC(x) → ∀xD(x)
LEQV
¬ ( ∀xA(x) ^ ∀xB(x) ^ ∀xC(x) ) v ∀xD(x) [→ Law]
LEQV
∃x¬A(x) v ∃x¬B(x) v ∃x¬C(x) v ∀xD(x) [Duality of quantifiers law]
LEQV
∃x1∃x2∃x3∀x ( ¬A(x1) v ¬B(x2) v ¬C(x3) v D(x) ) [pulling quantifiers out]
LEQV
∃x1∃x2∃x3∀x ( A(x1) → (¬B(x2) v ¬C(x3) v D(x) ) ) [reverse → Law ]
LEQV
∃x1∃x2∃x3∀x ( A(x1) →( B(x2) →( ¬C(x3) v D(x) )) ) [reverse → Law ]
LEQV
∃x1∃x2∃x3∀x ( A(x1) →( (B(x2) → (C(x3) → D(x) )) ) [reverse → Law ]
Re: Help with prenex normal form
→ v D(x)
is not syntactical.
I guess you mean:
Ax1-> (Bx2 -> (Cx3-> Dx))
Re: Help with prenex normal form
yeah, i edited the last post to fix it, but other than it, is each line logically equivilant?
Re: Help with prenex normal form
Darn, I messed up. I better doublecheck. I need to doublecheck the order of the quantifiers.
Re: Help with prenex normal form
Unless I'm missing something (and I'll use 'V' and 'E' as the quantifiers), we have:
(VyAy & VzBz & VwCw) -> VxDx
equivalent to
VxEyzw(Ay -> (Bz -> (Cw -> Dx)))
But I don't see how you would get
(VyAy & VzBz & VwCw) -> VxDx
equivalent to
EyzwVx(Ay -> (Bz -> (Cw -> Dx)))
I think I was correct when I said (before edits) that you reversed the correct order of quantifiers when you "pulled out".
Re: Help with prenex normal form
Now I am confused...
I thought I could do what I did in post#3, and since it ended up leqv to the last one, it has to be leqv.
Re: Help with prenex normal form
Look at your line where you pulled out the quantifiers. I think you incorrectly reversed the order.
Re: Help with prenex normal form
I don't see how this is wrong
∃x1∃x2∃x3∀x ( ¬A(x1) v ¬B(x2) v ¬C(x3) v D(x) ) [pulling quantifiers out]
because when I pull out the quantifiers, I renamed the variables using this law
Discrete Structures, Logic, and ... - Google Books
so each variable in the quantifiers are different, so the order shouldn't matter right?
Re: Help with prenex normal form
Suppose
and
are quantifiers (∀ or ∃) and
is ∧ or ∨. Then
)\star(\mathcal{Q}_2y\,B(y)))
is equivalent to both
\star B(y)))
and
\star B(y)))
(assuming x does not occur freely in B and y does not occur freely in A). Therefore, in this problem the order of quantifiers in the prenex form does not matter.
Re: Help with prenex normal form
Re: Help with prenex normal form
Re: Help with prenex normal form
Yeah, my mistake. Both are correct.