proving a logical consequence when subtracting a tautology from a set

Question

IfI have a set X of Well-formed formulae and a Formula P. I know that X ⊨ P.

If i am givin a tautology Q

then prove that X - {Q} ⊨ P

I don't understand why subtracting the set Q wouldn't make a difference to P being a logical consequence X.

Answer 1

We know that Q is a tautology already, right? So, for every valuation v that satisfies a set X, i.e. $v \vDash X$, then we see that $v \vDash X$ U $\top$. For this same reason, $\top$ is irrelevant when contained in a set of assumptions: we can always subtract it preserving logical consequences.

Answer 2

Let $v$ be a valuation that makes all formulas of $X\setminus \{Q\}$ true. Since $Q$ is a tautology, one gets $\forall \phi\in X(v(\phi)=1)$.

Answer 3

Another way of proving it is by basic properties of the tautological consequence relation : $\vDash$

Let $\Sigma$ be a set of formulae and $\varphi, \psi$ formulae; then :

• $\Sigma \cup \{ \varphi \} \vDash \psi \quad$ iff $\quad \Sigma \vDash (\varphi \rightarrow \psi)$ --- (*)

and :

• if $\Sigma \vDash \varphi \quad$ and $\quad \Sigma \vDash (\varphi → \psi)$, then $\quad \Sigma \vDash \psi$ --- (§).

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Consider now $\Gamma \vDash \beta$, and let $\Gamma' = \Gamma \backslash \{ \alpha \}$.

If $\alpha$ is a tautology, we have that : $\vDash \alpha$, and thus $\Sigma \vDash \alpha$ for any set $\Sigma$ of formulae; in particular :

i) $\quad \Gamma' \vDash \alpha$.

But $\Gamma \vDash \beta$ is $\Gamma' \cup \{ \alpha \} \vDash \beta$.

Thus, from (*) we have that :

ii) $\quad \Gamma' \vDash (\alpha \rightarrow \beta)$.

Now apply (§) to i) and ii) to conclude with :

$\Gamma' \vDash \beta$.