If F satisfiable then ¬F is unsatisfiable.
$begingroup$
If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.
[A]α = ¬(¬[A]))
= ¬([A])
= ¬ (0)
= 1
logic propositional-calculus first-order-logic satisfiability
$endgroup$
add a comment |
$begingroup$
If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.
[A]α = ¬(¬[A]))
= ¬([A])
= ¬ (0)
= 1
logic propositional-calculus first-order-logic satisfiability
$endgroup$
add a comment |
$begingroup$
If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.
[A]α = ¬(¬[A]))
= ¬([A])
= ¬ (0)
= 1
logic propositional-calculus first-order-logic satisfiability
$endgroup$
If F satisfiable then ¬F is unsatisfiable. I know this is false and to show this I need to show a contradiction, this is my attempted answer, any ideas where I'm going wrong, this is revision for an exam.
[A]α = ¬(¬[A]))
= ¬([A])
= ¬ (0)
= 1
logic propositional-calculus first-order-logic satisfiability
logic propositional-calculus first-order-logic satisfiability
asked Jan 16 at 15:55
OctavianOctavian
32
32
add a comment |
add a comment |
1 Answer
1
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$begingroup$
In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).
This does not implies that its negation is unsatisfiable.
Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.
Clearly $F$ is sat, and also $lnot F$ is sat.
But there are many more : $p lor q, p to q, p land q, ldots$
The interesting relation is :
if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).
$endgroup$
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).
This does not implies that its negation is unsatisfiable.
Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.
Clearly $F$ is sat, and also $lnot F$ is sat.
But there are many more : $p lor q, p to q, p land q, ldots$
The interesting relation is :
if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).
$endgroup$
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
add a comment |
$begingroup$
In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).
This does not implies that its negation is unsatisfiable.
Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.
Clearly $F$ is sat, and also $lnot F$ is sat.
But there are many more : $p lor q, p to q, p land q, ldots$
The interesting relation is :
if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).
$endgroup$
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
add a comment |
$begingroup$
In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).
This does not implies that its negation is unsatisfiable.
Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.
Clearly $F$ is sat, and also $lnot F$ is sat.
But there are many more : $p lor q, p to q, p land q, ldots$
The interesting relation is :
if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).
$endgroup$
In propositional logic, a formula $F$ is satisfiable when there is a valuation (or : truth-assignment) $v$ such that $v(F)=$ t (in first-order logic : when there is an interpretation that makes the formula true).
This does not implies that its negation is unsatisfiable.
Consider the simple example of a formula $F := p$, where $p$ is a sentential letter.
Clearly $F$ is sat, and also $lnot F$ is sat.
But there are many more : $p lor q, p to q, p land q, ldots$
The interesting relation is :
if a formula $F$ is unsatisfiable, then $lnot F$ is a tautology (i.e. always true).
edited Jan 18 at 19:41
answered Jan 16 at 16:14
Mauro ALLEGRANZAMauro ALLEGRANZA
65.8k449114
65.8k449114
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
add a comment |
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
what do you mean by that example p is a sentential letter?
$endgroup$
– Octavian
Jan 16 at 17:02
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
$begingroup$
@Octavian - it is the formula $p$ (made of one only symbol).
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 19:31
add a comment |
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