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
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 :