bnj1415

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	bnj1415.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
	bnj1415.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1415.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
	bnj1415.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
	bnj1415.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
	bnj1415.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
	bnj1415.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
	bnj1415.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
	bnj1415.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
	bnj1415.10	⊢ 𝑃 = ∪ 𝐻
Assertion	bnj1415	⊢ ( 𝜒 → dom 𝑃 = trCl ( 𝑥 , 𝐴 , 𝑅 ) )

Proof

Step	Hyp	Ref	Expression
1		bnj1415.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2		bnj1415.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3		bnj1415.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
4		bnj1415.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
5		bnj1415.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
6		bnj1415.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
7		bnj1415.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
8		bnj1415.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
9		bnj1415.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
10		bnj1415.10	⊢ 𝑃 = ∪ 𝐻
11	6	simplbi	⊢ ( 𝜓 → 𝑅 FrSe 𝐴 )
12	7 11	bnj835	⊢ ( 𝜒 → 𝑅 FrSe 𝐴 )
13	5 7	bnj1212	⊢ ( 𝜒 → 𝑥 ∈ 𝐴 )
14		eqid	⊢ ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) )
15	14	bnj1414	⊢ ( ( 𝑅 FrSe 𝐴 ∧ 𝑥 ∈ 𝐴 ) → trCl ( 𝑥 , 𝐴 , 𝑅 ) = ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) ) )
16	12 13 15	syl2anc	⊢ ( 𝜒 → trCl ( 𝑥 , 𝐴 , 𝑅 ) = ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) ) )
17		iunun	⊢ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = ( ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) { 𝑦 } ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) )
18		iunid	⊢ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) { 𝑦 } = pred ( 𝑥 , 𝐴 , 𝑅 )
19	18	uneq1i	⊢ ( ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) { 𝑦 } ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) )
20	17 19	eqtri	⊢ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) )
21		biid	⊢ ( ( 𝜒 ∧ 𝑧 ∈ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) ↔ ( 𝜒 ∧ 𝑧 ∈ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
22		biid	⊢ ( ( ( 𝜒 ∧ 𝑧 ∈ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) ∧ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ∧ 𝑧 ∈ ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) ↔ ( ( 𝜒 ∧ 𝑧 ∈ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) ∧ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ∧ 𝑧 ∈ ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
23	1 2 3 4 5 6 7 8 9 10 21 22	bnj1398	⊢ ( 𝜒 → ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = dom 𝑃 )
24	20 23	eqtr3id	⊢ ( 𝜒 → ( pred ( 𝑥 , 𝐴 , 𝑅 ) ∪ ∪ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) trCl ( 𝑦 , 𝐴 , 𝑅 ) ) = dom 𝑃 )
25	16 24	eqtr2d	⊢ ( 𝜒 → dom 𝑃 = trCl ( 𝑥 , 𝐴 , 𝑅 ) )

Metamath Proof Explorer

Theorem bnj1415

Proof