Prove formula using skolemization and resolution
I have the following formula that I am trying to prove is valid:
$$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$
By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:
$$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
$$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
$$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
$$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
$$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
$$text{(no change)} qquad qquad text{Distribute matrix}$$
$$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$
I'm not sure how to reach the empty clause from here. I'd appreciate a hint.
first-order-logic quantifiers
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I have the following formula that I am trying to prove is valid:
$$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$
By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:
$$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
$$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
$$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
$$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
$$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
$$text{(no change)} qquad qquad text{Distribute matrix}$$
$$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$
I'm not sure how to reach the empty clause from here. I'd appreciate a hint.
first-order-logic quantifiers
add a comment |
I have the following formula that I am trying to prove is valid:
$$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$
By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:
$$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
$$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
$$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
$$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
$$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
$$text{(no change)} qquad qquad text{Distribute matrix}$$
$$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$
I'm not sure how to reach the empty clause from here. I'd appreciate a hint.
first-order-logic quantifiers
I have the following formula that I am trying to prove is valid:
$$exists x forall y q(x,y) rightarrow forall y exists x q(x,y)$$
By following the Skolemization algorithm in the book "Mathematical Logic for Computer Science" by Mordechai Ben-Ari, I get the following:
$$neg (exists x forall y q(x,y) rightarrow forall y exists x q(x,y)) qquad qquad text{Negated formula}$$
$$neg (exists x forall y q(x,y) rightarrow forall w exists z q(z,w)) qquad qquad text{Rename bound variables}$$
$$neg (neg exists x forall y q(x,y) lor forall w exists z q(z,w)) qquad qquad text{Eliminate boolean operators}$$
$$exists x forall y q(x,y) land exists w forall z neg q(z,w) qquad qquad text{Push negation inwards}$$
$$exists x forall y exists w forall z (q(x,y) land neg q(z,w)) qquad qquad text{Extract quantifiers}$$
$$text{(no change)} qquad qquad text{Distribute matrix}$$
$$forall y forall z (q(a,y) land neg q(z,f(y))) qquad qquad text{Replace existential quantifiers}$$
I'm not sure how to reach the empty clause from here. I'd appreciate a hint.
first-order-logic quantifiers
first-order-logic quantifiers
edited Dec 4 '18 at 2:32
asked Dec 4 '18 at 2:26
Steve
1027
1027
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When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$
That is, you get:
$exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$
$exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$
$exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$
$exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$
$exists x exists w forall y forall z (q(x,y) land neg q(z,w))$
Skolemizing now gets you:
$forall y forall z (q(a,y) land neg q(z,b))$
Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$
And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.
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1 Answer
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1 Answer
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When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$
That is, you get:
$exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$
$exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$
$exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$
$exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$
$exists x exists w forall y forall z (q(x,y) land neg q(z,w))$
Skolemizing now gets you:
$forall y forall z (q(a,y) land neg q(z,b))$
Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$
And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.
add a comment |
When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$
That is, you get:
$exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$
$exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$
$exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$
$exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$
$exists x exists w forall y forall z (q(x,y) land neg q(z,w))$
Skolemizing now gets you:
$forall y forall z (q(a,y) land neg q(z,b))$
Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$
And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.
add a comment |
When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$
That is, you get:
$exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$
$exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$
$exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$
$exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$
$exists x exists w forall y forall z (q(x,y) land neg q(z,w))$
Skolemizing now gets you:
$forall y forall z (q(a,y) land neg q(z,b))$
Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$
And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.
When you extract the quantifiers, pull out the $exists w$ before you pull out the $forall y$... that'll eliminate the dependency of $w$ on $y$
That is, you get:
$exists x forall y q(x,y) land exists w forall z neg q(z,w) Leftrightarrow$
$exists x (forall y q(x,y) land exists w forall z neg q(z,w)) Leftrightarrow$
$exists x exists w (forall y q(x,y) land forall z neg q(z,w)) Leftrightarrow$
$exists x exists w forall y (q(x,y) land forall z neg q(z,w) Leftrightarrow$
$exists x exists w forall y forall z (q(x,y) land neg q(z,w))$
Skolemizing now gets you:
$forall y forall z (q(a,y) land neg q(z,b))$
Thus clauses ${q(a,y)}$ and ${neg q(z,b)}$
And by substituting $a$ for $z$ and $b$ for $y$ these can be resolved to the empty clause.
edited Dec 4 '18 at 3:16
answered Dec 4 '18 at 3:05
Bram28
60.3k44590
60.3k44590
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