First Order Logic: how to prove the formula?












1












$begingroup$


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







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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


















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






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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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12











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0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can we prove it using beginning of my proof? Is it possible?
    $endgroup$
    – Aleksandra
    Dec 10 '18 at 8:12














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 11:27









Mauro ALLEGRANZAMauro ALLEGRANZA

67.2k449115




67.2k449115












  • $begingroup$
    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




$begingroup$
Can we prove it using beginning of my proof? Is it possible?
$endgroup$
– Aleksandra
Dec 10 '18 at 8:12


















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