Finding prenex normal form of a formula
$begingroup$
Find prenex normal form of the formula $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
My attempt:
- $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
- $(exists x)S(x,y)rightarrow (R(x)rightarrow (forall u)neg S(x,u))$
- $(exists x)S(x,y)rightarrow (forall u)(R(x)rightarrow neg S(x,u))$
- $(forall u)((exists x)S(x,y)rightarrow (R(x)rightarrow neg S(x,u)))$
- $(forall u)(forall w)(S(w,y)rightarrow (R(x)rightarrow neg S(x,u)))$
I am wondering if the last step is correct. Can anybody tell?
logic first-order-logic predicate-logic quantifiers
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
$endgroup$
add a comment |
$begingroup$
Find prenex normal form of the formula $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
My attempt:
- $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
- $(exists x)S(x,y)rightarrow (R(x)rightarrow (forall u)neg S(x,u))$
- $(exists x)S(x,y)rightarrow (forall u)(R(x)rightarrow neg S(x,u))$
- $(forall u)((exists x)S(x,y)rightarrow (R(x)rightarrow neg S(x,u)))$
- $(forall u)(forall w)(S(w,y)rightarrow (R(x)rightarrow neg S(x,u)))$
I am wondering if the last step is correct. Can anybody tell?
logic first-order-logic predicate-logic quantifiers
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
$endgroup$
add a comment |
$begingroup$
Find prenex normal form of the formula $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
My attempt:
- $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
- $(exists x)S(x,y)rightarrow (R(x)rightarrow (forall u)neg S(x,u))$
- $(exists x)S(x,y)rightarrow (forall u)(R(x)rightarrow neg S(x,u))$
- $(forall u)((exists x)S(x,y)rightarrow (R(x)rightarrow neg S(x,u)))$
- $(forall u)(forall w)(S(w,y)rightarrow (R(x)rightarrow neg S(x,u)))$
I am wondering if the last step is correct. Can anybody tell?
logic first-order-logic predicate-logic quantifiers
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
$endgroup$
Find prenex normal form of the formula $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
My attempt:
- $(exists x)S(x,y)rightarrow (R(x)rightarrow neg(exists u)S(x,u))$
- $(exists x)S(x,y)rightarrow (R(x)rightarrow (forall u)neg S(x,u))$
- $(exists x)S(x,y)rightarrow (forall u)(R(x)rightarrow neg S(x,u))$
- $(forall u)((exists x)S(x,y)rightarrow (R(x)rightarrow neg S(x,u)))$
- $(forall u)(forall w)(S(w,y)rightarrow (R(x)rightarrow neg S(x,u)))$
I am wondering if the last step is correct. Can anybody tell?
logic first-order-logic predicate-logic quantifiers
logic first-order-logic predicate-logic quantifiers
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
edited Dec 24 '18 at 14:36
Bram28
63.3k44793
63.3k44793
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
asked Dec 24 '18 at 13:52
Leyla AlkanLeyla Alkan
1,5801724
1,5801724
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes, that is correct, though I would break that step into two: first replace the variable, and then bring out the quantifier. So:
$(forall u) ((exists x) S(x,y) rightarrow (R(x) rightarrow neg S(x,u))) overset{text{Replace variables}}Leftrightarrow$
$(forall u) ((exists w) S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))overset{text{Prenex Law}}Leftrightarrow$
$(forall u) (forall w) (S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))$
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, that is correct, though I would break that step into two: first replace the variable, and then bring out the quantifier. So:
$(forall u) ((exists x) S(x,y) rightarrow (R(x) rightarrow neg S(x,u))) overset{text{Replace variables}}Leftrightarrow$
$(forall u) ((exists w) S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))overset{text{Prenex Law}}Leftrightarrow$
$(forall u) (forall w) (S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))$
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
Yes, that is correct, though I would break that step into two: first replace the variable, and then bring out the quantifier. So:
$(forall u) ((exists x) S(x,y) rightarrow (R(x) rightarrow neg S(x,u))) overset{text{Replace variables}}Leftrightarrow$
$(forall u) ((exists w) S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))overset{text{Prenex Law}}Leftrightarrow$
$(forall u) (forall w) (S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))$
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
Yes, that is correct, though I would break that step into two: first replace the variable, and then bring out the quantifier. So:
$(forall u) ((exists x) S(x,y) rightarrow (R(x) rightarrow neg S(x,u))) overset{text{Replace variables}}Leftrightarrow$
$(forall u) ((exists w) S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))overset{text{Prenex Law}}Leftrightarrow$
$(forall u) (forall w) (S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))$
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
$endgroup$
Yes, that is correct, though I would break that step into two: first replace the variable, and then bring out the quantifier. So:
$(forall u) ((exists x) S(x,y) rightarrow (R(x) rightarrow neg S(x,u))) overset{text{Replace variables}}Leftrightarrow$
$(forall u) ((exists w) S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))overset{text{Prenex Law}}Leftrightarrow$
$(forall u) (forall w) (S(w,y) rightarrow (R(x) rightarrow neg S(x,y)))$
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
edited Dec 24 '18 at 14:13
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
answered Dec 24 '18 at 14:08
Bram28Bram28
63.3k44793
63.3k44793
add a comment |
add a comment |
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