Proof using natural deduction (Tautology)
$begingroup$
I've been asked to prove the following tautology via natural deduction:
$forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$
I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.
First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.
Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.
Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
Is there a way to make this third deduction or did I just start my whole proof wrong?
logic first-order-logic predicate-logic natural-deduction formal-proofs
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
$endgroup$
add a comment |
$begingroup$
I've been asked to prove the following tautology via natural deduction:
$forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$
I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.
First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.
Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.
Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
Is there a way to make this third deduction or did I just start my whole proof wrong?
logic first-order-logic predicate-logic natural-deduction formal-proofs
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
$endgroup$
add a comment |
$begingroup$
I've been asked to prove the following tautology via natural deduction:
$forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$
I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.
First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.
Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.
Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
Is there a way to make this third deduction or did I just start my whole proof wrong?
logic first-order-logic predicate-logic natural-deduction formal-proofs
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
$endgroup$
I've been asked to prove the following tautology via natural deduction:
$forall x , (lnot Px lor Qx) rightarrow (forall y , Py rightarrow forall z ,Qz)$
I normally use tree proofs, but I don't think I can show those here so I'll say in words what I've done so far.
First, I assume $forall x (lnot Px lor Qx)$ to deduce $lnot Pd lor Qd$ from $forall x , (lnot Px lor Qx)$.
Secondly, I assume $forall y , Py$ and $lnot Pd$ to deduce $(forall y , Py rightarrow forall , z Qz)$ from $lnot Pd$.
Thirdly, I assume $Qd$ and am trying to deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
If I can make this third deduction I can use OR-elimination to get the conclusion, but I don't see how I can deduce $(forall y , Py rightarrow forall z , Qz)$ from $Qd$.
Is there a way to make this third deduction or did I just start my whole proof wrong?
logic first-order-logic predicate-logic natural-deduction formal-proofs
logic first-order-logic predicate-logic natural-deduction formal-proofs
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
edited Jan 5 at 15:51
Taroccoesbrocco
5,77971840
5,77971840
asked Jan 5 at 13:43
NaborDHNaborDH
304
asked Jan 5 at 13:43
NaborDHNaborDH
304
asked Jan 5 at 13:43
NaborDHNaborDH
304
304
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).
Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.
begin{equation}
dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
end{equation}
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
$endgroup$
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).
Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.
begin{equation}
dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
end{equation}
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
$endgroup$
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
add a comment |
$begingroup$
To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).
Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.
begin{equation}
dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
end{equation}
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
$endgroup$
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
add a comment |
$begingroup$
To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).
Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.
begin{equation}
dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
end{equation}
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
$endgroup$
To deduce $forall y Py to forall z Qz$ from $Qd$, you should deduce $forall z Q z$ from the assumption $Qd$, but this is impossible because of the restriction on the free variable for the rule $forall_i$ (see here for a discussion of the issue).
Thus, the right approach is to apply the rule $lor_e$ in order to deduce $Qd$ without any assumption on $d$ (i.e. $Qd$ should be discharged), in this way you can correctly apply the rule $forall_i$ to deduce $forall z Q z$. Concretely, the following is a derivation in natural deduction of $forall x(lnot Px lor Qx) to (forall y Py to forall z Qz)$.
begin{equation}
dfrac{dfrac{[forall x (lnot Px lor Qx)]^circ}{lnot Pz lor Qz}forall_e qquad dfrac{dfrac{[lnot Pz]^* qquad dfrac{[forall y Py]^bullet}{Pz}forall_e}{bot}lnot_e}{Qz}text{efq} qquad [Qz]^*}{dfrac{Qz}{dfrac{forall z Qz}{dfrac{forall y Py to forall z Qz}{forall x (lnot Px lor Qx) to (forall y to forall z Qz)}to_i^circ}to_i^bullet}forall_i} lor_e^*
end{equation}
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
answered Jan 5 at 15:45
TaroccoesbroccoTaroccoesbrocco
5,77971840
5,77971840
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
add a comment |
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
Thanks for the detailed explanation.
$endgroup$
– NaborDH
Jan 5 at 17:18
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
$begingroup$
@NaborDH - You are welcome!
$endgroup$
– Taroccoesbrocco
Jan 5 at 19:03
add a comment |
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