bnj981

Description: Technical lemma for bnj69 . 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
Hypotheses	bnj981.1	⊢ ( 𝜑 ↔ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) )
	bnj981.2	⊢ ( 𝜓 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
	bnj981.3	⊢ 𝐷 = ( ω ∖ { ∅ } )
	bnj981.4	⊢ 𝐵 = { 𝑓 ∣ ∃ 𝑛 ∈ 𝐷 ( 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) }
	bnj981.5	⊢ ( 𝜒 ↔ ( 𝑛 ∈ 𝐷 ∧ 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) )
Assertion	bnj981	⊢ ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) )

Proof

Step	Hyp	Ref	Expression
1		bnj981.1	⊢ ( 𝜑 ↔ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) )
2		bnj981.2	⊢ ( 𝜓 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
3		bnj981.3	⊢ 𝐷 = ( ω ∖ { ∅ } )
4		bnj981.4	⊢ 𝐵 = { 𝑓 ∣ ∃ 𝑛 ∈ 𝐷 ( 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) }
5		bnj981.5	⊢ ( 𝜒 ↔ ( 𝑛 ∈ 𝐷 ∧ 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) )
6		nfv	⊢ Ⅎ 𝑦 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 )
7		nfcv	⊢ Ⅎ 𝑦 ω
8		nfv	⊢ Ⅎ 𝑦 suc 𝑖 ∈ 𝑛
9		nfiu1	⊢ Ⅎ 𝑦 ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
10	9	nfeq2	⊢ Ⅎ 𝑦 ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
11	8 10	nfim	⊢ Ⅎ 𝑦 ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
12	7 11	nfralw	⊢ Ⅎ 𝑦 ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
13	2 12	nfxfr	⊢ Ⅎ 𝑦 𝜓
14	13	nf5ri	⊢ ( 𝜓 → ∀ 𝑦 𝜓 )
15	14 5	bnj1096	⊢ ( 𝜒 → ∀ 𝑦 𝜒 )
16	15	nf5i	⊢ Ⅎ 𝑦 𝜒
17		nfv	⊢ Ⅎ 𝑦 𝑖 ∈ 𝑛
18		nfv	⊢ Ⅎ 𝑦 𝑍 ∈ ( 𝑓 ‘ 𝑖 )
19	16 17 18	nf3an	⊢ Ⅎ 𝑦 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) )
20	19	nfex	⊢ Ⅎ 𝑦 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) )
21	20	nfex	⊢ Ⅎ 𝑦 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) )
22	21	nfex	⊢ Ⅎ 𝑦 ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) )
23	6 22	nfim	⊢ Ⅎ 𝑦 ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) )
24		eleq1	⊢ ( 𝑦 = 𝑍 → ( 𝑦 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) ↔ 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) ) )
25		eleq1	⊢ ( 𝑦 = 𝑍 → ( 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) ↔ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) )
26	25	3anbi3d	⊢ ( 𝑦 = 𝑍 → ( ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) ) ↔ ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) ) )
27	26	3exbidv	⊢ ( 𝑦 = 𝑍 → ( ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) ) ↔ ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) ) )
28	24 27	imbi12d	⊢ ( 𝑦 = 𝑍 → ( ( 𝑦 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) ) ) ↔ ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) ) ) )
29	1 2 3 4 5	bnj917	⊢ ( 𝑦 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) ) )
30	23 28 29	vtoclg1f	⊢ ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) ) )
31	30	pm2.43i	⊢ ( 𝑍 ∈ trCl ( 𝑋 , 𝐴 , 𝑅 ) → ∃ 𝑓 ∃ 𝑛 ∃ 𝑖 ( 𝜒 ∧ 𝑖 ∈ 𝑛 ∧ 𝑍 ∈ ( 𝑓 ‘ 𝑖 ) ) )

Metamath Proof Explorer

Theorem bnj981

Proof