Metamath Proof Explorer


Theorem bnj1234

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 bnj1234.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
bnj1234.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
bnj1234.4 𝑍 = ⟨ 𝑥 , ( 𝑔 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
bnj1234.5 𝐷 = { 𝑔 ∣ ∃ 𝑑𝐵 ( 𝑔 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) }
Assertion bnj1234 𝐶 = 𝐷

Proof

Step Hyp Ref Expression
1 bnj1234.2 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
2 bnj1234.3 𝐶 = { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) }
3 bnj1234.4 𝑍 = ⟨ 𝑥 , ( 𝑔 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
4 bnj1234.5 𝐷 = { 𝑔 ∣ ∃ 𝑑𝐵 ( 𝑔 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) }
5 fneq1 ( 𝑓 = 𝑔 → ( 𝑓 Fn 𝑑𝑔 Fn 𝑑 ) )
6 fveq1 ( 𝑓 = 𝑔 → ( 𝑓𝑥 ) = ( 𝑔𝑥 ) )
7 reseq1 ( 𝑓 = 𝑔 → ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) = ( 𝑔 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) )
8 7 opeq2d ( 𝑓 = 𝑔 → ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩ = ⟨ 𝑥 , ( 𝑔 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩ )
9 8 1 3 3eqtr4g ( 𝑓 = 𝑔𝑌 = 𝑍 )
10 9 fveq2d ( 𝑓 = 𝑔 → ( 𝐺𝑌 ) = ( 𝐺𝑍 ) )
11 6 10 eqeq12d ( 𝑓 = 𝑔 → ( ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ↔ ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) )
12 11 ralbidv ( 𝑓 = 𝑔 → ( ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ↔ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) )
13 5 12 anbi12d ( 𝑓 = 𝑔 → ( ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ↔ ( 𝑔 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) ) )
14 13 rexbidv ( 𝑓 = 𝑔 → ( ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) ↔ ∃ 𝑑𝐵 ( 𝑔 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) ) )
15 14 cbvabv { 𝑓 ∣ ∃ 𝑑𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑓𝑥 ) = ( 𝐺𝑌 ) ) } = { 𝑔 ∣ ∃ 𝑑𝐵 ( 𝑔 Fn 𝑑 ∧ ∀ 𝑥𝑑 ( 𝑔𝑥 ) = ( 𝐺𝑍 ) ) }
16 15 2 4 3eqtr4i 𝐶 = 𝐷