Metamath Proof Explorer


Theorem bnj1514

Description: Technical lemma for bnj1500 . 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 bnj1514.1 𝐵 = { 𝑑 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
bnj1514.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
bnj1514.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
Assertion bnj1514 ( 𝑓𝐶 → ∀ 𝑥 ∈ dom 𝑓 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) )

Proof

Step Hyp Ref Expression
1 bnj1514.1 𝐵 = { 𝑑 ∣ ( 𝑑𝐴 ∧ ∀ 𝑥𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
2 bnj1514.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
3 bnj1514.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
4 3 bnj1436 ( 𝑓𝐶 → ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) )
5 df-rex ( ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ↔ ∃ 𝑑 ( 𝑑𝐵 ∧ ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ) )
6 3anass ( ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ↔ ( 𝑑𝐵 ∧ ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ) )
7 5 6 bnj133 ( ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ↔ ∃ 𝑑 ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) )
8 4 7 sylib ( 𝑓𝐶 → ∃ 𝑑 ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) )
9 simp3 ( ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) → ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) )
10 fndm ( 𝑓 Fn 𝑑 → dom 𝑓 = 𝑑 )
11 10 3ad2ant2 ( ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) → dom 𝑓 = 𝑑 )
12 11 raleqdv ( ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) → ( ∀ 𝑥 ∈ dom 𝑓 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ↔ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) )
13 9 12 mpbird ( ( 𝑑𝐵𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) → ∀ 𝑥 ∈ dom 𝑓 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) )
14 8 13 bnj593 ( 𝑓𝐶 → ∃ 𝑑𝑥 ∈ dom 𝑓 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) )
15 14 bnj937 ( 𝑓𝐶 → ∀ 𝑥 ∈ dom 𝑓 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) )