Metamath Proof Explorer


Theorem 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 ) ) )