Metamath Proof Explorer

Theorem frege75

Description: If from the proposition that x has property A , whatever x may be, it can be inferred that every result of an application of the procedure R to x has property A , then property A is hereditary in the R -sequence. Proposition 75 of Frege1879 p. 60. (Contributed by RP, 28-Mar-2020) (Proof modification is discouraged.)

		Ref	Expression
	Assertion	frege75	$⊢ \forall x (x \in A \to \forall y (x R y \to y \in A)) \to R hereditary A$

Proof

Step	Hyp	Ref	Expression
1		dffrege69	$⊢ \forall x (x \in A \to \forall y (x R y \to y \in A)) \leftrightarrow R hereditary A$
2		frege52aid	$⊢ (\forall x (x \in A \to \forall y (x R y \to y \in A)) \leftrightarrow R hereditary A) \to (\forall x (x \in A \to \forall y (x R y \to y \in A)) \to R hereditary A)$
3	1 2	ax-mp	$⊢ \forall x (x \in A \to \forall y (x R y \to y \in A)) \to R hereditary A$