Metamath Proof Explorer


Theorem fin23lem36

Description: Lemma for fin23 . Weak order property of Y . (Contributed by Stefan O'Rear, 2-Nov-2014)

Ref Expression
Hypotheses fin23lem33.f ⊢ F = g | ∀ a ∈ 𝒫 g ω ∀ x ∈ ω a ⁡ suc ⁡ x ⊆ a ⁡ x → ⋂ ran ⁡ a ∈ ran ⁡ a
fin23lem.f ⊢ φ → h : ω ⟶ 1-1 V
fin23lem.g ⊢ φ → ⋃ ran ⁡ h ⊆ G
fin23lem.h ⊢ φ → ∀ j j : ω ⟶ 1-1 V ∧ ⋃ ran ⁡ j ⊆ G → i ⁡ j : ω ⟶ 1-1 V ∧ ⋃ ran ⁡ i ⁡ j ⊂ ⋃ ran ⁡ j
fin23lem.i ⊢ Y = rec ⁡ i h ↾ ω
Assertion fin23lem36 ⊢ A ∈ ω ∧ B ∈ ω ∧ B ⊆ A ∧ φ → ⋃ ran ⁡ Y ⁡ A ⊆ ⋃ ran ⁡ Y ⁡ B

Proof

Step Hyp Ref Expression
1 fin23lem33.f ⊢ F = g | ∀ a ∈ 𝒫 g ω ∀ x ∈ ω a ⁡ suc ⁡ x ⊆ a ⁡ x → ⋂ ran ⁡ a ∈ ran ⁡ a
2 fin23lem.f ⊢ φ → h : ω ⟶ 1-1 V
3 fin23lem.g ⊢ φ → ⋃ ran ⁡ h ⊆ G
4 fin23lem.h ⊢ φ → ∀ j j : ω ⟶ 1-1 V ∧ ⋃ ran ⁡ j ⊆ G → i ⁡ j : ω ⟶ 1-1 V ∧ ⋃ ran ⁡ i ⁡ j ⊂ ⋃ ran ⁡ j
5 fin23lem.i ⊢ Y = rec ⁡ i h ↾ ω
6 fveq2 ⊢ a = B → Y ⁡ a = Y ⁡ B
7 6 rneqd ⊢ a = B → ran ⁡ Y ⁡ a = ran ⁡ Y ⁡ B
8 7 unieqd ⊢ a = B → ⋃ ran ⁡ Y ⁡ a = ⋃ ran ⁡ Y ⁡ B
9 8 sseq1d ⊢ a = B → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ ⋃ ran ⁡ Y ⁡ B ⊆ ⋃ ran ⁡ Y ⁡ B
10 9 imbi2d ⊢ a = B → φ → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ φ → ⋃ ran ⁡ Y ⁡ B ⊆ ⋃ ran ⁡ Y ⁡ B
11 fveq2 ⊢ a = b → Y ⁡ a = Y ⁡ b
12 11 rneqd ⊢ a = b → ran ⁡ Y ⁡ a = ran ⁡ Y ⁡ b
13 12 unieqd ⊢ a = b → ⋃ ran ⁡ Y ⁡ a = ⋃ ran ⁡ Y ⁡ b
14 13 sseq1d ⊢ a = b → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
15 14 imbi2d ⊢ a = b → φ → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ φ → ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
16 fveq2 ⊢ a = suc ⁡ b → Y ⁡ a = Y ⁡ suc ⁡ b
17 16 rneqd ⊢ a = suc ⁡ b → ran ⁡ Y ⁡ a = ran ⁡ Y ⁡ suc ⁡ b
18 17 unieqd ⊢ a = suc ⁡ b → ⋃ ran ⁡ Y ⁡ a = ⋃ ran ⁡ Y ⁡ suc ⁡ b
19 18 sseq1d ⊢ a = suc ⁡ b → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
20 19 imbi2d ⊢ a = suc ⁡ b → φ → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ φ → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
21 fveq2 ⊢ a = A → Y ⁡ a = Y ⁡ A
22 21 rneqd ⊢ a = A → ran ⁡ Y ⁡ a = ran ⁡ Y ⁡ A
23 22 unieqd ⊢ a = A → ⋃ ran ⁡ Y ⁡ a = ⋃ ran ⁡ Y ⁡ A
24 23 sseq1d ⊢ a = A → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ ⋃ ran ⁡ Y ⁡ A ⊆ ⋃ ran ⁡ Y ⁡ B
25 24 imbi2d ⊢ a = A → φ → ⋃ ran ⁡ Y ⁡ a ⊆ ⋃ ran ⁡ Y ⁡ B ↔ φ → ⋃ ran ⁡ Y ⁡ A ⊆ ⋃ ran ⁡ Y ⁡ B
26 ssid ⊢ ⋃ ran ⁡ Y ⁡ B ⊆ ⋃ ran ⁡ Y ⁡ B
27 26 2a1i ⊢ B ∈ ω → φ → ⋃ ran ⁡ Y ⁡ B ⊆ ⋃ ran ⁡ Y ⁡ B
28 simprr ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b ∧ φ → φ
29 simpll ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b ∧ φ → b ∈ ω
30 1 2 3 4 5 fin23lem35 ⊢ φ ∧ b ∈ ω → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊂ ⋃ ran ⁡ Y ⁡ b
31 28 29 30 syl2anc ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b ∧ φ → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊂ ⋃ ran ⁡ Y ⁡ b
32 31 pssssd ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b ∧ φ → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ b
33 sstr2 ⊢ ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ b → ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
34 32 33 syl ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b ∧ φ → ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
35 34 expr ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b → φ → ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
36 35 a2d ⊢ b ∈ ω ∧ B ∈ ω ∧ B ⊆ b → φ → ⋃ ran ⁡ Y ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B → φ → ⋃ ran ⁡ Y ⁡ suc ⁡ b ⊆ ⋃ ran ⁡ Y ⁡ B
37 10 15 20 25 27 36 findsg ⊢ A ∈ ω ∧ B ∈ ω ∧ B ⊆ A → φ → ⋃ ran ⁡ Y ⁡ A ⊆ ⋃ ran ⁡ Y ⁡ B
38 37 impr ⊢ A ∈ ω ∧ B ∈ ω ∧ B ⊆ A ∧ φ → ⋃ ran ⁡ Y ⁡ A ⊆ ⋃ ran ⁡ Y ⁡ B