Metamath Proof Explorer


Theorem f1opw2

Description: A one-to-one mapping induces a one-to-one mapping on power sets. This version of f1opw avoids the Axiom of Replacement. (Contributed by Mario Carneiro, 26-Jun-2015)

Ref Expression
Hypotheses f1opw2.1 ⊢ φ → F : A ⟶ 1-1 onto B
f1opw2.2 ⊢ φ → F -1 a ∈ V
f1opw2.3 ⊢ φ → F b ∈ V
Assertion f1opw2 ⊢ φ → b ∈ 𝒫 A ⟼ F b : 𝒫 A ⟶ 1-1 onto 𝒫 B

Proof

Step Hyp Ref Expression
1 f1opw2.1 ⊢ φ → F : A ⟶ 1-1 onto B
2 f1opw2.2 ⊢ φ → F -1 a ∈ V
3 f1opw2.3 ⊢ φ → F b ∈ V
4 eqid ⊢ b ∈ 𝒫 A ⟼ F b = b ∈ 𝒫 A ⟼ F b
5 imassrn ⊢ F b ⊆ ran ⁡ F
6 f1ofo ⊢ F : A ⟶ 1-1 onto B → F : A ⟶ onto B
7 1 6 syl ⊢ φ → F : A ⟶ onto B
8 forn ⊢ F : A ⟶ onto B → ran ⁡ F = B
9 7 8 syl ⊢ φ → ran ⁡ F = B
10 5 9 sseqtrid ⊢ φ → F b ⊆ B
11 3 10 elpwd ⊢ φ → F b ∈ 𝒫 B
12 11 adantr ⊢ φ ∧ b ∈ 𝒫 A → F b ∈ 𝒫 B
13 imassrn ⊢ F -1 a ⊆ ran ⁡ F -1
14 dfdm4 ⊢ dom ⁡ F = ran ⁡ F -1
15 f1odm ⊢ F : A ⟶ 1-1 onto B → dom ⁡ F = A
16 1 15 syl ⊢ φ → dom ⁡ F = A
17 14 16 eqtr3id ⊢ φ → ran ⁡ F -1 = A
18 13 17 sseqtrid ⊢ φ → F -1 a ⊆ A
19 2 18 elpwd ⊢ φ → F -1 a ∈ 𝒫 A
20 19 adantr ⊢ φ ∧ a ∈ 𝒫 B → F -1 a ∈ 𝒫 A
21 elpwi ⊢ a ∈ 𝒫 B → a ⊆ B
22 21 adantl ⊢ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → a ⊆ B
23 foimacnv ⊢ F : A ⟶ onto B ∧ a ⊆ B → F F -1 a = a
24 7 22 23 syl2an ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → F F -1 a = a
25 24 eqcomd ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → a = F F -1 a
26 imaeq2 ⊢ b = F -1 a → F b = F F -1 a
27 26 eqeq2d ⊢ b = F -1 a → a = F b ↔ a = F F -1 a
28 25 27 syl5ibrcom ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → b = F -1 a → a = F b
29 f1of1 ⊢ F : A ⟶ 1-1 onto B → F : A ⟶ 1-1 B
30 1 29 syl ⊢ φ → F : A ⟶ 1-1 B
31 elpwi ⊢ b ∈ 𝒫 A → b ⊆ A
32 31 adantr ⊢ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → b ⊆ A
33 f1imacnv ⊢ F : A ⟶ 1-1 B ∧ b ⊆ A → F -1 F b = b
34 30 32 33 syl2an ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → F -1 F b = b
35 34 eqcomd ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → b = F -1 F b
36 imaeq2 ⊢ a = F b → F -1 a = F -1 F b
37 36 eqeq2d ⊢ a = F b → b = F -1 a ↔ b = F -1 F b
38 35 37 syl5ibrcom ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → a = F b → b = F -1 a
39 28 38 impbid ⊢ φ ∧ b ∈ 𝒫 A ∧ a ∈ 𝒫 B → b = F -1 a ↔ a = F b
40 4 12 20 39 f1o2d ⊢ φ → b ∈ 𝒫 A ⟼ F b : 𝒫 A ⟶ 1-1 onto 𝒫 B