bnj590

Description: Technical lemma for bnj852 . This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

		Ref	Expression
	Hypothesis	bnj590.1	$⊢ ψ \leftrightarrow \forall i \in ω (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R))$
	Assertion	bnj590	$⊢ (B = suc i \land ψ) \to (i \in ω \to (B \in n \to f (B) = ⋃_{y \in f (i)} pred (y, A, R)))$

Proof

Step	Hyp	Ref	Expression
1		bnj590.1	$⊢ ψ \leftrightarrow \forall i \in ω (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R))$
2		rsp	$⊢ \forall i \in ω (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R)) \to (i \in ω \to (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R)))$
3	1 2	sylbi	$⊢ ψ \to (i \in ω \to (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R)))$
4		eleq1	$⊢ B = suc i \to (B \in n \leftrightarrow suc i \in n)$
5		fveqeq2	$⊢ B = suc i \to (f (B) = ⋃_{y \in f (i)} pred (y, A, R) \leftrightarrow f (suc i) = ⋃_{y \in f (i)} pred (y, A, R))$
6	4 5	imbi12d	$⊢ B = suc i \to ((B \in n \to f (B) = ⋃_{y \in f (i)} pred (y, A, R)) \leftrightarrow (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R)))$
7	6	imbi2d	$⊢ B = suc i \to ((i \in ω \to (B \in n \to f (B) = ⋃_{y \in f (i)} pred (y, A, R))) \leftrightarrow (i \in ω \to (suc i \in n \to f (suc i) = ⋃_{y \in f (i)} pred (y, A, R))))$
8	3 7	syl5ibr	$⊢ B = suc i \to (ψ \to (i \in ω \to (B \in n \to f (B) = ⋃_{y \in f (i)} pred (y, A, R))))$
9	8	imp	$⊢ (B = suc i \land ψ) \to (i \in ω \to (B \in n \to f (B) = ⋃_{y \in f (i)} pred (y, A, R)))$

Metamath Proof Explorer

Theorem bnj590

Proof