Metamath Proof Explorer


Theorem oacomf1olem

Description: Lemma for oacomf1o . (Contributed by Mario Carneiro, 30-May-2015)

Ref Expression
Hypothesis oacomf1olem.1 ⊢ F = x ∈ A ⟼ B + 𝑜 x
Assertion oacomf1olem ⊢ A ∈ On ∧ B ∈ On → F : A ⟶ 1-1 onto ran ⁡ F ∧ ran ⁡ F ∩ B = ∅

Proof

Step Hyp Ref Expression
1 oacomf1olem.1 ⊢ F = x ∈ A ⟼ B + 𝑜 x
2 oaf1o ⊢ B ∈ On → x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 onto On ∖ B
3 2 adantl ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 onto On ∖ B
4 f1of1 ⊢ x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 onto On ∖ B → x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 On ∖ B
5 3 4 syl ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 On ∖ B
6 onss ⊢ A ∈ On → A ⊆ On
7 6 adantr ⊢ A ∈ On ∧ B ∈ On → A ⊆ On
8 f1ssres ⊢ x ∈ On ⟼ B + 𝑜 x : On ⟶ 1-1 On ∖ B ∧ A ⊆ On → x ∈ On ⟼ B + 𝑜 x ↾ A : A ⟶ 1-1 On ∖ B
9 5 7 8 syl2anc ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x ↾ A : A ⟶ 1-1 On ∖ B
10 7 resmptd ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x ↾ A = x ∈ A ⟼ B + 𝑜 x
11 10 1 eqtr4di ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x ↾ A = F
12 f1eq1 ⊢ x ∈ On ⟼ B + 𝑜 x ↾ A = F → x ∈ On ⟼ B + 𝑜 x ↾ A : A ⟶ 1-1 On ∖ B ↔ F : A ⟶ 1-1 On ∖ B
13 11 12 syl ⊢ A ∈ On ∧ B ∈ On → x ∈ On ⟼ B + 𝑜 x ↾ A : A ⟶ 1-1 On ∖ B ↔ F : A ⟶ 1-1 On ∖ B
14 9 13 mpbid ⊢ A ∈ On ∧ B ∈ On → F : A ⟶ 1-1 On ∖ B
15 f1f1orn ⊢ F : A ⟶ 1-1 On ∖ B → F : A ⟶ 1-1 onto ran ⁡ F
16 14 15 syl ⊢ A ∈ On ∧ B ∈ On → F : A ⟶ 1-1 onto ran ⁡ F
17 f1f ⊢ F : A ⟶ 1-1 On ∖ B → F : A ⟶ On ∖ B
18 frn ⊢ F : A ⟶ On ∖ B → ran ⁡ F ⊆ On ∖ B
19 14 17 18 3syl ⊢ A ∈ On ∧ B ∈ On → ran ⁡ F ⊆ On ∖ B
20 19 difss2d ⊢ A ∈ On ∧ B ∈ On → ran ⁡ F ⊆ On
21 reldisj ⊢ ran ⁡ F ⊆ On → ran ⁡ F ∩ B = ∅ ↔ ran ⁡ F ⊆ On ∖ B
22 20 21 syl ⊢ A ∈ On ∧ B ∈ On → ran ⁡ F ∩ B = ∅ ↔ ran ⁡ F ⊆ On ∖ B
23 19 22 mpbird ⊢ A ∈ On ∧ B ∈ On → ran ⁡ F ∩ B = ∅
24 16 23 jca ⊢ A ∈ On ∧ B ∈ On → F : A ⟶ 1-1 onto ran ⁡ F ∧ ran ⁡ F ∩ B = ∅