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 ∣ ( 𝑑 ⊆ 𝐴 ∧ ∀ 𝑥 ∈ 𝑑 pred ( 𝑥 , 𝐴 , 𝑅 ) ⊆ 𝑑 ) }
	bnj1514.2	⊢ 𝑌 = ⟨ 𝑥 , ( 𝑓 ↾ pred ( 𝑥 , 𝐴 , 𝑅 ) ) ⟩
	bnj1514.3	⊢ 𝐶 = { 𝑓 ∣ ∃ 𝑑 ∈ 𝐵 ( 𝑓 Fn 𝑑 ∧ ∀ 𝑥 ∈ 𝑑 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) ) }
Assertion	bnj1514	⊢ ( 𝑓 ∈ 𝐶 → ∀ 𝑥 ∈ dom 𝑓 ( 𝑓 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑌 ) )

Proof

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

Metamath Proof Explorer

Theorem bnj1514

Proof