Is $ (\forall x)(A \rightarrow B \land C) \rightarrow (\forall x)(A \rightarrow B) $ an absolute theorem schema?

Question

Is $ (\forall x)(A \rightarrow B \land C) \rightarrow (\forall x)(A \rightarrow B) $ an absolute theorem schema ?

• If you think 'yes', then give a proof.
• If you think 'no', construct a counter model or prove the invalid strong generalization from it.

I wanna make sure if my answer is right.

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What I have tried :

Yes!

By deduction theorem , $ (\forall x)(A \rightarrow B \land C) \vdash (\forall x)(A \rightarrow B)$

(1) $(\forall x)(A \rightarrow B \land C) $ (hypothesis)

(2) $ A \rightarrow (B \land C) $ (1, spec)

(3) $\lnot A \lor (B \land C)$ (2, 2.4.11 + Eqn)

(4) $(\lnot A \lor B) \land (\lnot A \lor C)$ (3, 2,4,23(i) + Eqn)

(5) $(\lnot A \lor B)$ (4, split)

(6) $ A \rightarrow B$ (5, 2.4.11 + Eqn)

(7) $ (\forall x)(A \rightarrow B)$ (6 + Weak generaliztion , x dnof in $\Gamma$ )

See George Tourlakis, Mathematical Logic (2008) or this post for a lists of theorems and axioms.

Answer 1

Basically, DanulG's comment gives you the needed hint : the formula is valid; thus, by completeness is provable.

Your proof is sound.

Another proof can be "manufactured" starting from the fact that :

$\vDash_{TAUT} (A \rightarrow (B \land C)) \rightarrow (A \rightarrow B)$

[easy check with truth-table].

Then apply 4.2.1 Definition (Logical Axiom Schemata of Predicate Logic) [page 139] : Ax1 : partial generalization of tautologies are axioms, to have :

$\vdash (\forall x)[(A \rightarrow (B \land C)) \rightarrow (A \rightarrow B)]$.

Finally, apply Ax3 : all formulae of the form $(\forall x)(A \rightarrow B) \rightarrow ((\forall x)A \rightarrow (\forall x)B)$ are axioms, and modus ponens to get :

$\vdash (\forall x)(A \rightarrow (B \land C)) \rightarrow (\forall x)(A \rightarrow B)$.