Metamath Proof Explorer


Theorem 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 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
bnj1491.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
bnj1491.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
bnj1491.4 ( 𝜏 ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
bnj1491.5 𝐷 = { 𝑥𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
bnj1491.6 ( 𝜓 ↔ ( 𝑅 FrSe 𝐴𝐷 ≠ ∅ ) )
bnj1491.7 ( 𝜒 ↔ ( 𝜓𝑥𝐷 ∧ ∀ 𝑦𝐷 ¬ 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 , ( 𝐺𝑍 ) ⟩ } )
bnj1491.13 ( 𝜒 → ( 𝑄𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
Assertion bnj1491 ( ( 𝜒𝑄 ∈ V ) → ∃ 𝑓 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )

Proof

Step Hyp Ref Expression
1 bnj1491.1 𝐵 = { 𝑑 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2 bnj1491.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3 bnj1491.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
4 bnj1491.4 ( 𝜏 ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
5 bnj1491.5 𝐷 = { 𝑥𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
6 bnj1491.6 ( 𝜓 ↔ ( 𝑅 FrSe 𝐴𝐷 ≠ ∅ ) )
7 bnj1491.7 ( 𝜒 ↔ ( 𝜓𝑥𝐷 ∧ ∀ 𝑦𝐷 ¬ 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 , ( 𝐺𝑍 ) ⟩ } )
13 bnj1491.13 ( 𝜒 → ( 𝑄𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
14 1 2 3 4 5 6 7 8 9 10 11 12 bnj1466 ( 𝑤𝑄 → ∀ 𝑓 𝑤𝑄 )
15 14 nfcii 𝑓 𝑄
16 3 bnj1317 ( 𝑤𝐶 → ∀ 𝑓 𝑤𝐶 )
17 16 nfcii 𝑓 𝐶
18 15 17 nfel 𝑓 𝑄𝐶
19 15 nfdm 𝑓 dom 𝑄
20 19 nfeq1 𝑓 dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) )
21 18 20 nfan 𝑓 ( 𝑄𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) )
22 eleq1 ( 𝑓 = 𝑄 → ( 𝑓𝐶𝑄𝐶 ) )
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 → ( ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ↔ ( 𝑄𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ) )
26 15 21 25 spcegf ( 𝑄 ∈ V → ( ( 𝑄𝐶 ∧ dom 𝑄 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) → ∃ 𝑓 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) ) )
27 13 26 mpan9 ( ( 𝜒𝑄 ∈ V ) → ∃ 𝑓 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )