bnj1416

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	bnj1416.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
	bnj1416.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1416.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
	bnj1416.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
	bnj1416.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
	bnj1416.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
	bnj1416.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
	bnj1416.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
	bnj1416.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
	bnj1416.10	⊢ 𝑃 = ∪ 𝐻
	bnj1416.11	⊢ 𝑍 = ⟨ 𝑥 , ( 𝑃 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1416.12	⊢ 𝑄 = ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
	bnj1416.28	⊢ ( 𝜒 → dom 𝑃 = trCl ( 𝑥 , 𝐴 , 𝑅 ) )
Assertion	bnj1416	⊢ ( 𝜒 → dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) )

Proof

Step	Hyp	Ref	Expression
1		bnj1416.1	⊢ 𝐵 = { 𝑑 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2		bnj1416.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3		bnj1416.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
4		bnj1416.4	⊢ ( 𝜏 ↔ ( 𝑓 ∈ 𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
5		bnj1416.5	⊢ 𝐷 = { 𝑥 ∈ 𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
6		bnj1416.6	⊢ ( 𝜓 ↔ ( 𝑅 FrSe 𝐴 ∧ 𝐷 ≠ ∅ ) )
7		bnj1416.7	⊢ ( 𝜒 ↔ ( 𝜓 ∧ 𝑥 ∈ 𝐷 ∧ ∀ 𝑦 ∈ 𝐷 ¬ 𝑦 𝑅 𝑥 ) )
8		bnj1416.8	⊢ ( 𝜏′ ↔ [ 𝑦 / 𝑥 ] 𝜏 )
9		bnj1416.9	⊢ 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
10		bnj1416.10	⊢ 𝑃 = ∪ 𝐻
11		bnj1416.11	⊢ 𝑍 = ⟨ 𝑥 , ( 𝑃 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
12		bnj1416.12	⊢ 𝑄 = ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
13		bnj1416.28	⊢ ( 𝜒 → dom 𝑃 = trCl ( 𝑥 , 𝐴 , 𝑅 ) )
14	12	dmeqi	⊢ dom 𝑄 = dom ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
15		dmun	⊢ dom ( 𝑃 ∪ { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } ) = ( dom 𝑃 ∪ dom { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } )
16		fvex	⊢ ( 𝐺 ‘ 𝑍 ) ∈ V
17	16	dmsnop	⊢ dom { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } = { 𝑥 }
18	17	uneq2i	⊢ ( dom 𝑃 ∪ dom { ⟨ 𝑥 , ( 𝐺 ‘ 𝑍 ) ⟩ } ) = ( dom 𝑃 ∪ { 𝑥 } )
19	14 15 18	3eqtri	⊢ dom 𝑄 = ( dom 𝑃 ∪ { 𝑥 } )
20	13	uneq1d	⊢ ( 𝜒 → ( dom 𝑃 ∪ { 𝑥 } ) = ( trCl ( 𝑥 , 𝐴 , 𝑅 ) ∪ { 𝑥 } ) )
21		uncom	⊢ ( trCl ( 𝑥 , 𝐴 , 𝑅 ) ∪ { 𝑥 } ) = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) )
22	20 21	eqtrdi	⊢ ( 𝜒 → ( dom 𝑃 ∪ { 𝑥 } ) = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) )
23	19 22	eqtrid	⊢ ( 𝜒 → dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) )

Metamath Proof Explorer

Theorem bnj1416

Proof