valid or Invalid

Question

$ \forall x (P(x) \rightarrow Q(x)) \rightarrow [\exists P(x)] \rightarrow [\forall Q(x)]$

 

how to check that a statement is valid or invalid ?

Comments

Invalid??

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yes but what is the approach?

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can you edit your question?

there is some ambiguity here?

$\rightarrow$ is Right associative (or) Left associative?

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is it like  $ T \rightarrow F$  try to put condition on LHS for F then for the same condition see the RHS of the IMplication

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Yes

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DOES implication have precedence and associativity involved with it ?

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Yes

$\rightarrow$ is have Precedence, but it is not associative,i think.

But $P\rightarrow Q\rightarrow R$ what is the answer?

if you take $\rightarrow$ is Left associative,then $(P\rightarrow Q)\rightarrow R$

(or)

if you take $\rightarrow$ is Right associative,then $P\rightarrow (Q\rightarrow R)$

Both cases answer might be different.

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→  has right associativity

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when operand surrounds  an operator of equal precedence the operand associate to  the right one .

this statement is written in stanford coursera course

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@ MiNiPanda and @hitendra singh

can you provide me resource??

I think this is not a rule?

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Ref: https://en.wikipedia.org/wiki/Material_conditional

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can you show how it is invalid ? I am facing difficulty in realizing that.

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if we take the condition as 

  P(x₁)=T                          Q(x₁).....Q(x_n)= T

P(x₂).....P(x_n-1)=F              Q(x_n)=F

P(x_n)=T

we can prove it invalid

Am I correct ?

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@hitendra singh

Can you please elaborate..I didn't get it..

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the solution is bit lengthy, but you can put

$\forall xP(x)=P_{1}.P_{2} $ and $\exists xP(x)=P_{1}+P_{2}$

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0

if we take the condition as 

  P(x₁)=T                          Q(x₁).....Q(x_n)= T

P(x₂).....P(x_n-1)=F              Q(x_n)=F

P(x_n)=T

 if both have n variables P and Q 

then 

[∃P(x)]→[∀Q(x)]   becomes false via above condition 

and ∀x(P(x)→Q(x))  becomes TRUE via above condition  thus over all  statement becomes false  

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Argument is valid iff there is no condition that premises are true and Conclusion are false. that means no condtion like T$\rightarrow$F should be there.

but if we put $\forall$Q(x)=F and $\forall$P(x)=T then full statement becomes false i.e. invalid

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@Dharmendra Lodhi 

Is my assumption right in order to prove that the statement is invalid ?

please look into it 

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@hitendra Singh

where you get this question?

it looks like an ambiguous question

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how is it  ambiguous ?

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$\rightarrow$ is no right associative??

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but its written on wikipedia

https://en.wikipedia.org/wiki/Material_conditional