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Given a theory $$ T= \{¬p ∨ ¬q\}\ over \ P = \{p, q\} $$ and theory $$ T' =\{p→r∧q, (q→r)→¬q\}\ over \ P' = \{p,q,r\} $$
I want to find if:

a) Is T' an extension of T?
b) Is T' conservative extension of T?

My attempt was the following:

a) T' is an extension of T since $ M(T')=\{\{0,0,1\},\{0,1,1\}\} $ and $ M(T)=\{\{0,1\},\{1,0\},\{0,0\}\} $, by definition T' is an extension of T iff all models of T' are models of T meaning $ M(T') ⊆ M(T) $ and also $ P ⊆ P' $. Since both of those hold we can say T' is an extension of T.

b) By definition T' is a conservative extension of T if $M(T)$ can be expanded to $M(T')$. So let's expand T by adding formulae $ φ = p → q ∧ p $ and let's call such theory $T*$ so $ T* = \{¬p ∨ ¬q, p → q ∧ p\} $ then we get that $M(T*)=M(T)$ and thus T' is conservative extension of T.

Is my reasoning correct? And if so can I add something else here? I have to say I am not sure about any of my answers since for some reason extensions seem to confuse me. Any help and useful tricks(if there is any) to quickly identify if T' is extension of T or not as well as simply find an extension of theory T will be appreciated.

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    a) looks right. b) is wrong: a model of $T$ that interprets $p$ as $\mathsf{true}$ cannot be expanded to a model of $T'$ (since in a model of $T'$, if $p$ is $\mathsf{true}$, then so is $q$ and this is not compatible with $T$). – 2017-02-09
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    @RobArthan thank you, just one more question, so what if I take $ T'' = \{ ¬p∨¬q, p→q \} $ and so $ M(T'') = \{ \{0,1\}, \{0,0\}\} $ . Do I understand this right that $ T'' $ will not be a conservative extension of $ T $ for the same reason as before ((1,0) cannot be expanded to $ T'' $ ) but now if I have another theories for example $ H = \{ p∨q\} $ and $ H' = \{ p∨q, p→q∨r \} $ then $ H'$ will be conservative extension of $H$ ? – 2017-02-09
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    That's right: the extension that makes $r$ $\mathsf{true}$ will always make $p \to q \lor r$ $\mathsf{true}$ . – 2017-02-09
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    @RobArthan Thank you, I think I got it now – 2017-02-09

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