Metamath Proof Explorer


Theorem domdifsn

Description: Dominance over a set with one element removed. (Contributed by Stefan O'Rear, 19-Feb-2015) (Revised by Mario Carneiro, 24-Jun-2015)

Ref Expression
Assertion domdifsn ⊢ A ≺ B → A ≼ B ∖ C

Proof

Step Hyp Ref Expression
1 sdomdom ⊢ A ≺ B → A ≼ B
2 relsdom ⊢ Rel ⁡ ≺
3 2 brrelex2i ⊢ A ≺ B → B ∈ V
4 brdomg ⊢ B ∈ V → A ≼ B ↔ ∃ f f : A ⟶ 1-1 B
5 3 4 syl ⊢ A ≺ B → A ≼ B ↔ ∃ f f : A ⟶ 1-1 B
6 1 5 mpbid ⊢ A ≺ B → ∃ f f : A ⟶ 1-1 B
7 6 adantr ⊢ A ≺ B ∧ C ∈ B → ∃ f f : A ⟶ 1-1 B
8 f1f ⊢ f : A ⟶ 1-1 B → f : A ⟶ B
9 8 frnd ⊢ f : A ⟶ 1-1 B → ran ⁡ f ⊆ B
10 9 adantl ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ran ⁡ f ⊆ B
11 sdomnen ⊢ A ≺ B → ¬ A ≈ B
12 11 ad2antrr ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ¬ A ≈ B
13 vex ⊢ f ∈ V
14 dff1o5 ⊢ f : A ⟶ 1-1 onto B ↔ f : A ⟶ 1-1 B ∧ ran ⁡ f = B
15 14 biimpri ⊢ f : A ⟶ 1-1 B ∧ ran ⁡ f = B → f : A ⟶ 1-1 onto B
16 f1oen3g ⊢ f ∈ V ∧ f : A ⟶ 1-1 onto B → A ≈ B
17 13 15 16 sylancr ⊢ f : A ⟶ 1-1 B ∧ ran ⁡ f = B → A ≈ B
18 17 ex ⊢ f : A ⟶ 1-1 B → ran ⁡ f = B → A ≈ B
19 18 necon3bd ⊢ f : A ⟶ 1-1 B → ¬ A ≈ B → ran ⁡ f ≠ B
20 19 adantl ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ¬ A ≈ B → ran ⁡ f ≠ B
21 12 20 mpd ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ran ⁡ f ≠ B
22 pssdifn0 ⊢ ran ⁡ f ⊆ B ∧ ran ⁡ f ≠ B → B ∖ ran ⁡ f ≠ ∅
23 10 21 22 syl2anc ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → B ∖ ran ⁡ f ≠ ∅
24 n0 ⊢ B ∖ ran ⁡ f ≠ ∅ ↔ ∃ x x ∈ B ∖ ran ⁡ f
25 23 24 sylib ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ∃ x x ∈ B ∖ ran ⁡ f
26 2 brrelex1i ⊢ A ≺ B → A ∈ V
27 26 ad2antrr ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → A ∈ V
28 3 ad2antrr ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → B ∈ V
29 28 difexd ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → B ∖ x ∈ V
30 eldifn ⊢ x ∈ B ∖ ran ⁡ f → ¬ x ∈ ran ⁡ f
31 disjsn ⊢ ran ⁡ f ∩ x = ∅ ↔ ¬ x ∈ ran ⁡ f
32 30 31 sylibr ⊢ x ∈ B ∖ ran ⁡ f → ran ⁡ f ∩ x = ∅
33 32 adantl ⊢ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → ran ⁡ f ∩ x = ∅
34 9 adantr ⊢ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → ran ⁡ f ⊆ B
35 reldisj ⊢ ran ⁡ f ⊆ B → ran ⁡ f ∩ x = ∅ ↔ ran ⁡ f ⊆ B ∖ x
36 34 35 syl ⊢ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → ran ⁡ f ∩ x = ∅ ↔ ran ⁡ f ⊆ B ∖ x
37 33 36 mpbid ⊢ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → ran ⁡ f ⊆ B ∖ x
38 f1ssr ⊢ f : A ⟶ 1-1 B ∧ ran ⁡ f ⊆ B ∖ x → f : A ⟶ 1-1 B ∖ x
39 37 38 syldan ⊢ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → f : A ⟶ 1-1 B ∖ x
40 39 adantl ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → f : A ⟶ 1-1 B ∖ x
41 f1dom2g ⊢ A ∈ V ∧ B ∖ x ∈ V ∧ f : A ⟶ 1-1 B ∖ x → A ≼ B ∖ x
42 27 29 40 41 syl3anc ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → A ≼ B ∖ x
43 eldifi ⊢ x ∈ B ∖ ran ⁡ f → x ∈ B
44 43 ad2antll ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → x ∈ B
45 simplr ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → C ∈ B
46 difsnen ⊢ B ∈ V ∧ x ∈ B ∧ C ∈ B → B ∖ x ≈ B ∖ C
47 28 44 45 46 syl3anc ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → B ∖ x ≈ B ∖ C
48 domentr ⊢ A ≼ B ∖ x ∧ B ∖ x ≈ B ∖ C → A ≼ B ∖ C
49 42 47 48 syl2anc ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B ∧ x ∈ B ∖ ran ⁡ f → A ≼ B ∖ C
50 49 expr ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → x ∈ B ∖ ran ⁡ f → A ≼ B ∖ C
51 50 exlimdv ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → ∃ x x ∈ B ∖ ran ⁡ f → A ≼ B ∖ C
52 25 51 mpd ⊢ A ≺ B ∧ C ∈ B ∧ f : A ⟶ 1-1 B → A ≼ B ∖ C
53 7 52 exlimddv ⊢ A ≺ B ∧ C ∈ B → A ≼ B ∖ C
54 1 adantr ⊢ A ≺ B ∧ ¬ C ∈ B → A ≼ B
55 difsn ⊢ ¬ C ∈ B → B ∖ C = B
56 55 breq2d ⊢ ¬ C ∈ B → A ≼ B ∖ C ↔ A ≼ B
57 56 adantl ⊢ A ≺ B ∧ ¬ C ∈ B → A ≼ B ∖ C ↔ A ≼ B
58 54 57 mpbird ⊢ A ≺ B ∧ ¬ C ∈ B → A ≼ B ∖ C
59 53 58 pm2.61dan ⊢ A ≺ B → A ≼ B ∖ C