Closure under tautological implication? Suppose $\Gamma$ is a consistent and complete set of formulas of sentential logic (built up from sentence letters using the sentential connectives). Suppose we define a new set $\Gamma^*$ to be the complement of $\Gamma$. (1) Is $\Gamma$ complete? (2) Is $\Gamma$ consistent? (3) Is it closed under tautological implication?
 A: The fact $\Gamma$ is a complete and consistent set of sentences does not imply that its complement $\Gamma^*$ is consistent or complete.


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*Let $\Gamma$ be any complete and consistent set of sentences (e.g. the set axioms of Presburger arithmetic) plus any universally valid sentence $A$ (e.g. the sentence $0=0$). Clearly, $\Gamma$ is complete and consistent. However, $\Gamma^*$ is not consistent because $\lnot A \in \Gamma^*$ by definition of complement, so $\Gamma^* \vdash \lnot A$, but also $\Gamma^* \vdash A$ by definition of validity. The error in your attempt of proof is that the fact that $\varphi \notin \Gamma^*$ does not imply that $\Gamma^* \not\vdash \varphi$. 

*Let $L$ be the language consisting of only an unary predicate symbol $P$ and a constant symbol $c$ (no identity). Let $\Gamma = \{\forall x P(x)\}$ (in $L$). Clearly, $\Gamma$ is complete and consistent, but its complement $\Gamma^* = \{\exists x \lnot P(x)\}$ is not complete because $\Gamma^* \not\vdash P(c)$ (take a model of $\Gamma^*$ with just one element) and $\Gamma^* \not\vdash \lnot P(c)$ (take a model of $\Gamma^*$ with two elements, the one that is the interpretation of $c$ has the property $P$ and the other one does not). The error in your attempt of proof is explained in Mauro Allegranza's comment.
