bnj1491

Description: Technical lemma for bnj60 . 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	bnj1491.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
	bnj1491.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1491.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
	bnj1491.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
	bnj1491.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
	bnj1491.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
	bnj1491.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
	bnj1491.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
	bnj1491.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
	bnj1491.10	⊢ 𝑃 = ∪ 𝐻
	bnj1491.11	⊢ 𝑍 = ⟨ 𝑥 , ( 𝑃 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1491.12	⊢ 𝑄 = ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
	bnj1491.13	⊢ ( 𝜒 → ( 𝑄 ∈ 𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
Assertion	bnj1491	⊢ ( ( 𝜒 ∧ 𝑄 ∈ V ) → ∃ 𝑓 ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		bnj1491.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2		bnj1491.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3		bnj1491.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
4		bnj1491.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
5		bnj1491.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
6		bnj1491.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
7		bnj1491.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
8		bnj1491.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
9		bnj1491.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
10		bnj1491.10	⊢ 𝑃 = ∪ 𝐻
11		bnj1491.11	⊢ 𝑍 = ⟨ 𝑥 , ( 𝑃 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
12		bnj1491.12	⊢ 𝑄 = ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
13		bnj1491.13	⊢ ( 𝜒 → ( 𝑄 ∈ 𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
14	1 2 3 4 5 6 7 8 9 10 11 12	bnj1466	⊢ ( 𝑤 ∈ 𝑄 → ∀ 𝑓 𝑤 ∈ 𝑄 )
15	14	nfcii	⊢ Ⅎ 𝑓 𝑄
16	3	bnj1317	⊢ ( 𝑤 ∈ 𝐶 → ∀ 𝑓 𝑤 ∈ 𝐶 )
17	16	nfcii	⊢ Ⅎ 𝑓 𝐶
18	15 17	nfel	⊢ Ⅎ 𝑓 𝑄 ∈ 𝐶
19	15	nfdm	⊢ Ⅎ 𝑓 dom 𝑄
20	19	nfeq1	⊢ Ⅎ 𝑓 dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) )
21	18 20	nfan	⊢ Ⅎ 𝑓 ( 𝑄 ∈ 𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) )
22		eleq1	⊢ ( 𝑓 = 𝑄 → ( 𝑓 ∈ 𝐶 ↔ 𝑄 ∈ 𝐶 ) )
23		dmeq	⊢ ( 𝑓 = 𝑄 → dom 𝑓 = dom 𝑄 )
24	23	eqeq1d	⊢ ( 𝑓 = 𝑄 → ( dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ↔ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
25	22 24	anbi12d	⊢ ( 𝑓 = 𝑄 → ( ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ↔ ( 𝑄 ∈ 𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ) )
26	15 21 25	spcegf	⊢ ( 𝑄 ∈ V → ( ( 𝑄 ∈ 𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) → ∃ 𝑓 ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ) )
27	13 26	mpan9	⊢ ( ( 𝜒 ∧ 𝑄 ∈ V ) → ∃ 𝑓 ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )

Metamath Proof Explorer

Theorem bnj1491

Proof