First Order Logic: how to prove the formula?
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How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?
$textbf{My work}:$
1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality
2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom
3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)
4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$
5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.
I dont know what to do next...
logic first-order-logic
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
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add a comment |
$begingroup$
How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?
$textbf{My work}:$
1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality
2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom
3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)
4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$
5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.
I dont know what to do next...
logic first-order-logic
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
$endgroup$
$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53
add a comment |
$begingroup$
How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?
$textbf{My work}:$
1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality
2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom
3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)
4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$
5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.
I dont know what to do next...
logic first-order-logic
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
$endgroup$
How to show this formula using axioms of First Order Logic: $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )rightarrow (P(x_0,x_2) rightarrow Q(x_0,x_2)))$?
$textbf{My work}:$
1) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom of equality
2) $forall x_{0}, forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)) rightarrowforall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - axiom
3)$forall x_{1}, forall x_2 ((x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1))$ - MP(1,2)
4) 2 times I repeat similar steps like in 1), 2), 3) and get $(x_{0}=x_{1})wedge P(x_1,x_2) rightarrow P(x_0,x_1)$
5) With $Q$ I do the same thing and get $(x_{0}=x_{1})wedge Q(x_1,x_2) rightarrow Q(x_0,x_1)$.
I dont know what to do next...
logic first-order-logic
logic first-order-logic
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
asked Dec 9 '18 at 9:12
Aleksandra Aleksandra
545
545
$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53
add a comment |
$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53
$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53
$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53
add a comment |
1 Answer
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Hint
The formula :
$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$
is a simple "variant" of the substitution axiom for equality :
$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$
where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.
Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.
Then apply the above axiom schema to get :
$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.
Finally, we have to use the Deduction Theorem followed by Generalization thrice.
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
$endgroup$
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Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint
The formula :
$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$
is a simple "variant" of the substitution axiom for equality :
$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$
where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.
Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.
Then apply the above axiom schema to get :
$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.
Finally, we have to use the Deduction Theorem followed by Generalization thrice.
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
$endgroup$
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
add a comment |
$begingroup$
Hint
The formula :
$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$
is a simple "variant" of the substitution axiom for equality :
$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$
where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.
Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.
Then apply the above axiom schema to get :
$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.
Finally, we have to use the Deduction Theorem followed by Generalization thrice.
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
$endgroup$
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
add a comment |
$begingroup$
Hint
The formula :
$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$
is a simple "variant" of the substitution axiom for equality :
$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$
where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.
Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.
Then apply the above axiom schema to get :
$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.
Finally, we have to use the Deduction Theorem followed by Generalization thrice.
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
$endgroup$
Hint
The formula :
$∀x_0,∀x_1,∀x_2((x_0=x_1) ∧ (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$
is a simple "variant" of the substitution axiom for equality :
$vdash ( x = y) to (phi[z:=x] to phi[z:=y])$
where $phi(z)$ is : $P(z,x_2) → Q(z,x_2)$.
Thus, we have to start from $(x_0=x_1)$ and with equality axioms derive : $(x_1=x_0)$.
Then apply the above axiom schema to get :
$(x_0=x_1) vdash (P(x_1,x_2) → Q(x_1,x_2)) → (P(x_0,x_2) → Q(x_0,x_2)))$.
Finally, we have to use the Deduction Theorem followed by Generalization thrice.
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
answered Dec 9 '18 at 11:27
Mauro ALLEGRANZAMauro ALLEGRANZA
66.2k449114
66.2k449114
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
add a comment |
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12
add a comment |
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$begingroup$
I thought about it $ ((x_{0}=x_{1})wedge (P(x_1,x_2) rightarrow Q(x_1,x_2) )vdash(P(x_0,x_2) rightarrow Q(x_0,x_2))$?? But I have no idea how to show it...
$endgroup$
– Aleksandra
Dec 9 '18 at 10:53