Metamath Proof Explorer


Theorem canthp1lem1

Description: Lemma for canthp1 . (Contributed by Mario Carneiro, 18-May-2015)

Ref Expression
Assertion canthp1lem1 ⊢ 1 𝑜 ≺ A → A ⊔︀ 2 𝑜 ≼ 𝒫 A

Proof

Step Hyp Ref Expression
1 1sdom2 ⊢ 1 𝑜 ≺ 2 𝑜
2 djuxpdom ⊢ 1 𝑜 ≺ A ∧ 1 𝑜 ≺ 2 𝑜 → A ⊔︀ 2 𝑜 ≼ A × 2 𝑜
3 1 2 mpan2 ⊢ 1 𝑜 ≺ A → A ⊔︀ 2 𝑜 ≼ A × 2 𝑜
4 sdom0 ⊢ ¬ 1 𝑜 ≺ ∅
5 breq2 ⊢ A = ∅ → 1 𝑜 ≺ A ↔ 1 𝑜 ≺ ∅
6 4 5 mtbiri ⊢ A = ∅ → ¬ 1 𝑜 ≺ A
7 6 con2i ⊢ 1 𝑜 ≺ A → ¬ A = ∅
8 neq0 ⊢ ¬ A = ∅ ↔ ∃ x x ∈ A
9 7 8 sylib ⊢ 1 𝑜 ≺ A → ∃ x x ∈ A
10 relsdom ⊢ Rel ⁡ ≺
11 10 brrelex2i ⊢ 1 𝑜 ≺ A → A ∈ V
12 11 adantr ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∈ V
13 enrefg ⊢ A ∈ V → A ≈ A
14 12 13 syl ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ≈ A
15 df2o2 ⊢ 2 𝑜 = ∅ ∅
16 pwpw0 ⊢ 𝒫 ∅ = ∅ ∅
17 15 16 eqtr4i ⊢ 2 𝑜 = 𝒫 ∅
18 0ex ⊢ ∅ ∈ V
19 vex ⊢ x ∈ V
20 en2sn ⊢ ∅ ∈ V ∧ x ∈ V → ∅ ≈ x
21 18 19 20 mp2an ⊢ ∅ ≈ x
22 pwen ⊢ ∅ ≈ x → 𝒫 ∅ ≈ 𝒫 x
23 21 22 ax-mp ⊢ 𝒫 ∅ ≈ 𝒫 x
24 17 23 eqbrtri ⊢ 2 𝑜 ≈ 𝒫 x
25 xpen ⊢ A ≈ A ∧ 2 𝑜 ≈ 𝒫 x → A × 2 𝑜 ≈ A × 𝒫 x
26 14 24 25 sylancl ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A × 2 𝑜 ≈ A × 𝒫 x
27 vsnex ⊢ x ∈ V
28 27 pwex ⊢ 𝒫 x ∈ V
29 uncom ⊢ A ∖ x ∪ x = x ∪ A ∖ x
30 simpr ⊢ 1 𝑜 ≺ A ∧ x ∈ A → x ∈ A
31 30 snssd ⊢ 1 𝑜 ≺ A ∧ x ∈ A → x ⊆ A
32 undif ⊢ x ⊆ A ↔ x ∪ A ∖ x = A
33 31 32 sylib ⊢ 1 𝑜 ≺ A ∧ x ∈ A → x ∪ A ∖ x = A
34 29 33 eqtrid ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ∪ x = A
35 12 difexd ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ∈ V
36 canth2g ⊢ A ∖ x ∈ V → A ∖ x ≺ 𝒫 A ∖ x
37 domunsn ⊢ A ∖ x ≺ 𝒫 A ∖ x → A ∖ x ∪ x ≼ 𝒫 A ∖ x
38 35 36 37 3syl ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ∪ x ≼ 𝒫 A ∖ x
39 34 38 eqbrtrrd ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ≼ 𝒫 A ∖ x
40 xpdom1g ⊢ 𝒫 x ∈ V ∧ A ≼ 𝒫 A ∖ x → A × 𝒫 x ≼ 𝒫 A ∖ x × 𝒫 x
41 28 39 40 sylancr ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A × 𝒫 x ≼ 𝒫 A ∖ x × 𝒫 x
42 endomtr ⊢ A × 2 𝑜 ≈ A × 𝒫 x ∧ A × 𝒫 x ≼ 𝒫 A ∖ x × 𝒫 x → A × 2 𝑜 ≼ 𝒫 A ∖ x × 𝒫 x
43 26 41 42 syl2anc ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A × 2 𝑜 ≼ 𝒫 A ∖ x × 𝒫 x
44 pwdjuen ⊢ A ∖ x ∈ V ∧ x ∈ V → 𝒫 A ∖ x ⊔︀ x ≈ 𝒫 A ∖ x × 𝒫 x
45 35 27 44 sylancl ⊢ 1 𝑜 ≺ A ∧ x ∈ A → 𝒫 A ∖ x ⊔︀ x ≈ 𝒫 A ∖ x × 𝒫 x
46 45 ensymd ⊢ 1 𝑜 ≺ A ∧ x ∈ A → 𝒫 A ∖ x × 𝒫 x ≈ 𝒫 A ∖ x ⊔︀ x
47 domentr ⊢ A × 2 𝑜 ≼ 𝒫 A ∖ x × 𝒫 x ∧ 𝒫 A ∖ x × 𝒫 x ≈ 𝒫 A ∖ x ⊔︀ x → A × 2 𝑜 ≼ 𝒫 A ∖ x ⊔︀ x
48 43 46 47 syl2anc ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A × 2 𝑜 ≼ 𝒫 A ∖ x ⊔︀ x
49 27 a1i ⊢ 1 𝑜 ≺ A ∧ x ∈ A → x ∈ V
50 disjdifr ⊢ A ∖ x ∩ x = ∅
51 50 a1i ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ∩ x = ∅
52 endjudisj ⊢ A ∖ x ∈ V ∧ x ∈ V ∧ A ∖ x ∩ x = ∅ → A ∖ x ⊔︀ x ≈ A ∖ x ∪ x
53 35 49 51 52 syl3anc ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ⊔︀ x ≈ A ∖ x ∪ x
54 53 34 breqtrd ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A ∖ x ⊔︀ x ≈ A
55 pwen ⊢ A ∖ x ⊔︀ x ≈ A → 𝒫 A ∖ x ⊔︀ x ≈ 𝒫 A
56 54 55 syl ⊢ 1 𝑜 ≺ A ∧ x ∈ A → 𝒫 A ∖ x ⊔︀ x ≈ 𝒫 A
57 domentr ⊢ A × 2 𝑜 ≼ 𝒫 A ∖ x ⊔︀ x ∧ 𝒫 A ∖ x ⊔︀ x ≈ 𝒫 A → A × 2 𝑜 ≼ 𝒫 A
58 48 56 57 syl2anc ⊢ 1 𝑜 ≺ A ∧ x ∈ A → A × 2 𝑜 ≼ 𝒫 A
59 9 58 exlimddv ⊢ 1 𝑜 ≺ A → A × 2 𝑜 ≼ 𝒫 A
60 domtr ⊢ A ⊔︀ 2 𝑜 ≼ A × 2 𝑜 ∧ A × 2 𝑜 ≼ 𝒫 A → A ⊔︀ 2 𝑜 ≼ 𝒫 A
61 3 59 60 syl2anc ⊢ 1 𝑜 ≺ A → A ⊔︀ 2 𝑜 ≼ 𝒫 A