Metamath Proof Explorer


Theorem domssex

Description: Weakening of domssex2 to forget the functions in favor of dominance and equinumerosity. (Contributed by Mario Carneiro, 7-Feb-2015) (Revised by Mario Carneiro, 24-Jun-2015)

Ref Expression
Assertion domssex ⊢ A ≼ B → ∃ x A ⊆ x ∧ B ≈ x

Proof

Step Hyp Ref Expression
1 brdomi ⊢ A ≼ B → ∃ f f : A ⟶ 1-1 B
2 reldom ⊢ Rel ⁡ ≼
3 2 brrelex2i ⊢ A ≼ B → B ∈ V
4 vex ⊢ f ∈ V
5 f1stres ⊢ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A : B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ⟶ B ∖ ran ⁡ f
6 difexg ⊢ B ∈ V → B ∖ ran ⁡ f ∈ V
7 6 adantl ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → B ∖ ran ⁡ f ∈ V
8 snex ⊢ 𝒫 ⋃ ran ⁡ A ∈ V
9 xpexg ⊢ B ∖ ran ⁡ f ∈ V ∧ 𝒫 ⋃ ran ⁡ A ∈ V → B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
10 7 8 9 sylancl ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
11 fex2 ⊢ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A : B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ⟶ B ∖ ran ⁡ f ∧ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V ∧ B ∖ ran ⁡ f ∈ V → 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
12 5 10 7 11 mp3an2i ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
13 unexg ⊢ f ∈ V ∧ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
14 4 12 13 sylancr ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V
15 cnvexg ⊢ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V
16 14 15 syl ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V
17 rnexg ⊢ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V → ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V
18 16 17 syl ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V
19 simpl ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → f : A ⟶ 1-1 B
20 f1dm ⊢ f : A ⟶ 1-1 B → dom ⁡ f = A
21 4 dmex ⊢ dom ⁡ f ∈ V
22 20 21 eqeltrrdi ⊢ f : A ⟶ 1-1 B → A ∈ V
23 22 adantr ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → A ∈ V
24 simpr ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → B ∈ V
25 eqid ⊢ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 = f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
26 25 domss2 ⊢ f : A ⟶ 1-1 B ∧ A ∈ V ∧ B ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 : B ⟶ 1-1 onto ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∘ f = I ↾ A
27 19 23 24 26 syl3anc ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 : B ⟶ 1-1 onto ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∘ f = I ↾ A
28 27 simp2d ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
29 27 simp1d ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 : B ⟶ 1-1 onto ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
30 f1oen3g ⊢ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∈ V ∧ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 : B ⟶ 1-1 onto ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 → B ≈ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
31 16 29 30 syl2anc ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → B ≈ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
32 28 31 jca ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ B ≈ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
33 sseq2 ⊢ x = ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 → A ⊆ x ↔ A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
34 breq2 ⊢ x = ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 → B ≈ x ↔ B ≈ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
35 33 34 anbi12d ⊢ x = ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 → A ⊆ x ∧ B ≈ x ↔ A ⊆ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1 ∧ B ≈ ran ⁡ f ∪ 1 st ↾ B ∖ ran ⁡ f × 𝒫 ⋃ ran ⁡ A -1
36 18 32 35 spcedv ⊢ f : A ⟶ 1-1 B ∧ B ∈ V → ∃ x A ⊆ x ∧ B ≈ x
37 36 ex ⊢ f : A ⟶ 1-1 B → B ∈ V → ∃ x A ⊆ x ∧ B ≈ x
38 37 exlimiv ⊢ ∃ f f : A ⟶ 1-1 B → B ∈ V → ∃ x A ⊆ x ∧ B ≈ x
39 1 3 38 sylc ⊢ A ≼ B → ∃ x A ⊆ x ∧ B ≈ x