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If $T_{1}\subset T_{2}$ and $T_{1}$ is complete and $T_{2}$ is satisfiable and both are in the same language , then $T_{1}= T_{2}$

I know a set is complete if $\forall y, y\in T$ or $\neg y\in T$. But I don't know how to apply this towards an actual proof. Can anyone help?

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    $\begingroup$ HINT: $T_1$ is a maximal consistent theory. $T_2$ is consistent, since it has a model. $\endgroup$ – Brian M. Scott Dec 5 '16 at 6:56
  • $\begingroup$ So you must mean $T_1 \subseteq T_2$ $\endgroup$ – Bram28 Dec 6 '16 at 3:38
  • $\begingroup$ @Bram28 what would be the difference ? $\endgroup$ – A curious one Dec 6 '16 at 3:48
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    $\begingroup$ $A \subseteq B$ means that $A$ is a subset of $B$, meaning that all elements of $A$ are in $B$. Notice that with this definition, we also have that $A \subseteq A$ for any set $A$: every set is a subset of itself. On the other hand, $A \subset B$ means that $A$ is a strict subset of $B$, meaning that (as with any subset) every element of $A$ is in $B$, but there are some elements in $B$ that are not in $A$. So, we never have that $A \subset A$. In your question we are to conclude that $T_1 = T_2$, which is impossible if $T_1 \subset T_2$. $\endgroup$ – Bram28 Dec 6 '16 at 12:46

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