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Is $ (\forall x)(A \rightarrow B) \vdash (\forall x)A \rightarrow (\forall x)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.

How can I do this one ?

Out Of Bounds
  • 1,531

1 Answers1

1

The inference is valid; here is a proof with Natural Deduction :

1) $(∀x)(A \rightarrow B)$ --- premise

2) $(∀x)A$ --- assumed [a]

3) $A \rightarrow B$ --- from 1) by $\forall$-elimination

4) $A$ --- from 2) by $\forall$-elimination

5) $B$ --- from 4) and 3) by $\rightarrow$-elimination

6) $\forall x B$ --- from 5) by $\forall$-introduction

7) $(∀x)A \rightarrow (∀x)B$ --- from 2) and 6) by $\rightarrow$-introduction, discharging assumption [a]

Thus, from 1) and 7) we have :

$(∀x)(A \rightarrow B) \vdash (∀x)A \rightarrow (∀x)B$.

Mauro ALLEGRANZA
  • 94,169