rankr1c

Description: A relationship between the rank function and the cumulative hierarchy of sets function R1 . Proposition 9.15(2) of TakeutiZaring p. 79. (Contributed by Mario Carneiro, 22-Mar-2013) (Revised by Mario Carneiro, 17-Nov-2014)

		Ref	Expression
	Assertion	rankr1c	$⊢ A \in ⋃ R_{1} [On] \to (B = rank (A) \leftrightarrow (\neg A \in R_{1} (B) \land A \in R_{1} (suc B)))$

Proof

Step	Hyp	Ref	Expression
1		id	$⊢ B = rank (A) \to B = rank (A)$
2		rankdmr1	$⊢ rank (A) \in dom R_{1}$
3	1 2	eqeltrdi	$⊢ B = rank (A) \to B \in dom R_{1}$
4	3	a1i	$⊢ A \in ⋃ R_{1} [On] \to (B = rank (A) \to B \in dom R_{1})$
5		elfvdm	$⊢ A \in R_{1} (suc B) \to suc B \in dom R_{1}$
6		r1dmlim	$⊢ Lim dom R_{1}$
7		limsuc	$⊢ Lim dom R_{1} \to (B \in dom R_{1} \leftrightarrow suc B \in dom R_{1})$
8	6 7	ax-mp	$⊢ B \in dom R_{1} \leftrightarrow suc B \in dom R_{1}$
9	5 8	sylibr	$⊢ A \in R_{1} (suc B) \to B \in dom R_{1}$
10	9	adantl	$⊢ (\neg A \in R_{1} (B) \land A \in R_{1} (suc B)) \to B \in dom R_{1}$
11	10	a1i	$⊢ A \in ⋃ R_{1} [On] \to ((\neg A \in R_{1} (B) \land A \in R_{1} (suc B)) \to B \in dom R_{1})$
12		eqss	$⊢ B = rank (A) \leftrightarrow (B \subseteq rank (A) \land rank (A) \subseteq B)$
13		rankr1clem	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to (\neg A \in R_{1} (B) \leftrightarrow B \subseteq rank (A))$
14		rankr1ag	$⊢ (A \in ⋃ R_{1} [On] \land suc B \in dom R_{1}) \to (A \in R_{1} (suc B) \leftrightarrow rank (A) \in suc B)$
15	8 14	sylan2b	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to (A \in R_{1} (suc B) \leftrightarrow rank (A) \in suc B)$
16		rankon	$⊢ rank (A) \in On$
17		limord	$⊢ Lim dom R_{1} \to Ord dom R_{1}$
18	6 17	ax-mp	$⊢ Ord dom R_{1}$
19		ordelon	$⊢ (Ord dom R_{1} \land B \in dom R_{1}) \to B \in On$
20	18 19	mpan	$⊢ B \in dom R_{1} \to B \in On$
21	20	adantl	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to B \in On$
22		onsssuc	$⊢ (rank (A) \in On \land B \in On) \to (rank (A) \subseteq B \leftrightarrow rank (A) \in suc B)$
23	16 21 22	sylancr	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to (rank (A) \subseteq B \leftrightarrow rank (A) \in suc B)$
24	15 23	bitr4d	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to (A \in R_{1} (suc B) \leftrightarrow rank (A) \subseteq B)$
25	13 24	anbi12d	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to ((\neg A \in R_{1} (B) \land A \in R_{1} (suc B)) \leftrightarrow (B \subseteq rank (A) \land rank (A) \subseteq B))$
26	12 25	bitr4id	$⊢ (A \in ⋃ R_{1} [On] \land B \in dom R_{1}) \to (B = rank (A) \leftrightarrow (\neg A \in R_{1} (B) \land A \in R_{1} (suc B)))$
27	26	ex	$⊢ A \in ⋃ R_{1} [On] \to (B \in dom R_{1} \to (B = rank (A) \leftrightarrow (\neg A \in R_{1} (B) \land A \in R_{1} (suc B))))$
28	4 11 27	pm5.21ndd	$⊢ A \in ⋃ R_{1} [On] \to (B = rank (A) \leftrightarrow (\neg A \in R_{1} (B) \land A \in R_{1} (suc B)))$

Metamath Proof Explorer

Theorem rankr1c

Proof