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Consider a set $ S $ of Boolean statements.

A set is said to be inconsistent if there does not exist an interpretation that makes all the statements true at the same time.

An interpretation being an assignment of "True"s and "False"'s to all the atoms of the statements that make up the set.

Question

If I remove a statement from $S$ that happens to a valid statement (i.e. a tautology) is $S$ still inconsistent?

I do not feel satisfied with the answer provided by my lecturer in the notes. Here it is.

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Maybe I am too dumb to understand it.

$\color{blue}{\textrm{Yes, If there does not exist an interpretation that makes all statements true at the same time}}$ $\color{red}{\textrm{and $a$ is valid}}$ $\color{blue}{\textrm{then there does not exist an interpretation that makes all statements true at the same time}}$

He just says that the fact that $a$ is valid (in conjunction with the fact that the set is inconsistent) implies that there does not exist an interpretation that makes all statements true at the same time. Why does the validity of $a$ imply this?

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  • $\begingroup$ What do you think is the answer? And why? $\endgroup$ – Asaf Karagila Dec 13 '19 at 23:14
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The proof is probably a little easier if you take the contrapositive:

Suppose that $S\setminus \{a\}$ is consistent. Then there is an interpretation $v$ that sets all statements in $S \setminus \{ a \}$ to True. But since $a$ is valid, $v$ will set $a$ to True as well. Hence, $v$ sets all statements in $S$ to True, and hence $S$ is consistent.

So, by contraposition, if $S$ is inconsistent, then $ S\setminus \{ a \}$ is inconsistent as well.

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