Metamath Proof Explorer


Theorem dju1dif

Description: Adding and subtracting one gives back the original cardinality. Similar to pncan for cardinalities. (Contributed by Mario Carneiro, 18-May-2015) (Revised by Jim Kingdon, 20-Aug-2023)

Ref Expression
Assertion dju1dif ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∖ B ≈ A

Proof

Step Hyp Ref Expression
1 simpl ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ∈ V
2 1oex ⊢ 1 𝑜 ∈ V
3 djuex ⊢ A ∈ V ∧ 1 𝑜 ∈ V → A ⊔︀ 1 𝑜 ∈ V
4 1 2 3 sylancl ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∈ V
5 simpr ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → B ∈ A ⊔︀ 1 𝑜
6 df1o2 ⊢ 1 𝑜 = ∅
7 6 xpeq2i ⊢ 1 𝑜 × 1 𝑜 = 1 𝑜 × ∅
8 0ex ⊢ ∅ ∈ V
9 2 8 xpsn ⊢ 1 𝑜 × ∅ = 1 𝑜 ∅
10 7 9 eqtri ⊢ 1 𝑜 × 1 𝑜 = 1 𝑜 ∅
11 ssun2 ⊢ 1 𝑜 × 1 𝑜 ⊆ ∅ × A ∪ 1 𝑜 × 1 𝑜
12 10 11 eqsstrri ⊢ 1 𝑜 ∅ ⊆ ∅ × A ∪ 1 𝑜 × 1 𝑜
13 opex ⊢ 1 𝑜 ∅ ∈ V
14 13 snss ⊢ 1 𝑜 ∅ ∈ ∅ × A ∪ 1 𝑜 × 1 𝑜 ↔ 1 𝑜 ∅ ⊆ ∅ × A ∪ 1 𝑜 × 1 𝑜
15 12 14 mpbir ⊢ 1 𝑜 ∅ ∈ ∅ × A ∪ 1 𝑜 × 1 𝑜
16 df-dju ⊢ A ⊔︀ 1 𝑜 = ∅ × A ∪ 1 𝑜 × 1 𝑜
17 15 16 eleqtrri ⊢ 1 𝑜 ∅ ∈ A ⊔︀ 1 𝑜
18 17 a1i ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → 1 𝑜 ∅ ∈ A ⊔︀ 1 𝑜
19 difsnen ⊢ A ⊔︀ 1 𝑜 ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 ∧ 1 𝑜 ∅ ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∖ B ≈ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅
20 4 5 18 19 syl3anc ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∖ B ≈ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅
21 16 difeq1i ⊢ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅ = ∅ × A ∪ 1 𝑜 × 1 𝑜 ∖ 1 𝑜 ∅
22 xp01disjl ⊢ ∅ × A ∩ 1 𝑜 × 1 𝑜 = ∅
23 disj3 ⊢ ∅ × A ∩ 1 𝑜 × 1 𝑜 = ∅ ↔ ∅ × A = ∅ × A ∖ 1 𝑜 × 1 𝑜
24 22 23 mpbi ⊢ ∅ × A = ∅ × A ∖ 1 𝑜 × 1 𝑜
25 difun2 ⊢ ∅ × A ∪ 1 𝑜 × 1 𝑜 ∖ 1 𝑜 × 1 𝑜 = ∅ × A ∖ 1 𝑜 × 1 𝑜
26 10 difeq2i ⊢ ∅ × A ∪ 1 𝑜 × 1 𝑜 ∖ 1 𝑜 × 1 𝑜 = ∅ × A ∪ 1 𝑜 × 1 𝑜 ∖ 1 𝑜 ∅
27 24 25 26 3eqtr2i ⊢ ∅ × A = ∅ × A ∪ 1 𝑜 × 1 𝑜 ∖ 1 𝑜 ∅
28 21 27 eqtr4i ⊢ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅ = ∅ × A
29 xpsnen2g ⊢ ∅ ∈ V ∧ A ∈ V → ∅ × A ≈ A
30 8 1 29 sylancr ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → ∅ × A ≈ A
31 28 30 eqbrtrid ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅ ≈ A
32 entr ⊢ A ⊔︀ 1 𝑜 ∖ B ≈ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅ ∧ A ⊔︀ 1 𝑜 ∖ 1 𝑜 ∅ ≈ A → A ⊔︀ 1 𝑜 ∖ B ≈ A
33 20 31 32 syl2anc ⊢ A ∈ V ∧ B ∈ A ⊔︀ 1 𝑜 → A ⊔︀ 1 𝑜 ∖ B ≈ A