Metamath Proof Explorer


Theorem bnj1371

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 bnj1371.1 𝐵 = { 𝑑 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
bnj1371.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
bnj1371.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
bnj1371.4 ( 𝜏 ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
bnj1371.5 𝐷 = { 𝑥𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
bnj1371.6 ( 𝜓 ↔ ( 𝑅 FrSe 𝐴𝐷 ≠ ∅ ) )
bnj1371.7 ( 𝜒 ↔ ( 𝜓𝑥𝐷 ∧ ∀ 𝑦𝐷 ¬ 𝑦 𝑅 𝑥 ) )
bnj1371.8 ( 𝜏′[ 𝑦 / 𝑥 ] 𝜏 )
bnj1371.9 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
bnj1371.10 𝑃 = 𝐻
bnj1371.11 ( 𝜏′ ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
Assertion bnj1371 ( 𝑓𝐻 → Fun 𝑓 )

Proof

Step Hyp Ref Expression
1 bnj1371.1 𝐵 = { 𝑑 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2 bnj1371.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3 bnj1371.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
4 bnj1371.4 ( 𝜏 ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑥 } ∪ trCl ( 𝑥 , 𝐴 , 𝑅 ) ) ) )
5 bnj1371.5 𝐷 = { 𝑥𝐴 ∣ ¬ ∃ 𝑓 𝜏 }
6 bnj1371.6 ( 𝜓 ↔ ( 𝑅 FrSe 𝐴𝐷 ≠ ∅ ) )
7 bnj1371.7 ( 𝜒 ↔ ( 𝜓𝑥𝐷 ∧ ∀ 𝑦𝐷 ¬ 𝑦 𝑅 𝑥 ) )
8 bnj1371.8 ( 𝜏′[ 𝑦 / 𝑥 ] 𝜏 )
9 bnj1371.9 𝐻 = { 𝑓 ∣ ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ }
10 bnj1371.10 𝑃 = 𝐻
11 bnj1371.11 ( 𝜏′ ↔ ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
12 9 bnj1436 ( 𝑓𝐻 → ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ )
13 rexex ( ∃ 𝑦 ∈ pred ( 𝑥 , 𝐴 , 𝑅 ) 𝜏′ → ∃ 𝑦 𝜏′ )
14 12 13 syl ( 𝑓𝐻 → ∃ 𝑦 𝜏′ )
15 11 exbii ( ∃ 𝑦 𝜏′ ↔ ∃ 𝑦 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
16 14 15 sylib ( 𝑓𝐻 → ∃ 𝑦 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
17 exsimpl ( ∃ 𝑦 ( 𝑓𝐶 ∧ dom 𝑓 = ( { 𝑦 } ∪ trCl ( 𝑦 , 𝐴 , 𝑅 ) ) ) → ∃ 𝑦 𝑓𝐶 )
18 16 17 syl ( 𝑓𝐻 → ∃ 𝑦 𝑓𝐶 )
19 3 abeq2i ( 𝑓𝐶 ↔ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) )
20 19 bnj1238 ( 𝑓𝐶 → ∃ 𝑑𝐵 𝑓 Fn 𝑑 )
21 fnfun ( 𝑓 Fn 𝑑 → Fun 𝑓 )
22 20 21 bnj31 ( 𝑓𝐶 → ∃ 𝑑𝐵 Fun 𝑓 )
23 22 bnj1265 ( 𝑓𝐶 → Fun 𝑓 )
24 18 23 bnj593 ( 𝑓𝐻 → ∃ 𝑦 Fun 𝑓 )
25 24 bnj937 ( 𝑓𝐻 → Fun 𝑓 )