Metamath Proof Explorer


Theorem canth

Description: No set A is equinumerous to its power set (Cantor's theorem), i.e., no function can map A onto its power set. Compare Theorem 6B(b) of Enderton p. 132. For the equinumerosity version, see canth2 . Note that A must be a set: this theorem does not hold when A is too large to be a set; see ncanth for a counterexample. (Use nex if you want the form -. E. f f : A -onto-> ~P A .) (Contributed by NM, 7-Aug-1994) (Proof shortened by Mario Carneiro, 7-Jun-2016)

Ref Expression
Hypothesis canth.1 ⊢ A ∈ V
Assertion canth ⊢ ¬ F : A ⟶ onto 𝒫 A

Proof

Step Hyp Ref Expression
1 canth.1 ⊢ A ∈ V
2 ssrab2 ⊢ x ∈ A | ¬ x ∈ F ⁡ x ⊆ A
3 1 2 elpwi2 ⊢ x ∈ A | ¬ x ∈ F ⁡ x ∈ 𝒫 A
4 forn ⊢ F : A ⟶ onto 𝒫 A → ran ⁡ F = 𝒫 A
5 3 4 eleqtrrid ⊢ F : A ⟶ onto 𝒫 A → x ∈ A | ¬ x ∈ F ⁡ x ∈ ran ⁡ F
6 id ⊢ x = y → x = y
7 fveq2 ⊢ x = y → F ⁡ x = F ⁡ y
8 6 7 eleq12d ⊢ x = y → x ∈ F ⁡ x ↔ y ∈ F ⁡ y
9 8 notbid ⊢ x = y → ¬ x ∈ F ⁡ x ↔ ¬ y ∈ F ⁡ y
10 9 elrab ⊢ y ∈ x ∈ A | ¬ x ∈ F ⁡ x ↔ y ∈ A ∧ ¬ y ∈ F ⁡ y
11 10 baibr ⊢ y ∈ A → ¬ y ∈ F ⁡ y ↔ y ∈ x ∈ A | ¬ x ∈ F ⁡ x
12 nbbn ⊢ ¬ y ∈ F ⁡ y ↔ y ∈ x ∈ A | ¬ x ∈ F ⁡ x ↔ ¬ y ∈ F ⁡ y ↔ y ∈ x ∈ A | ¬ x ∈ F ⁡ x
13 11 12 sylib ⊢ y ∈ A → ¬ y ∈ F ⁡ y ↔ y ∈ x ∈ A | ¬ x ∈ F ⁡ x
14 eleq2 ⊢ F ⁡ y = x ∈ A | ¬ x ∈ F ⁡ x → y ∈ F ⁡ y ↔ y ∈ x ∈ A | ¬ x ∈ F ⁡ x
15 13 14 nsyl ⊢ y ∈ A → ¬ F ⁡ y = x ∈ A | ¬ x ∈ F ⁡ x
16 15 nrex ⊢ ¬ ∃ y ∈ A F ⁡ y = x ∈ A | ¬ x ∈ F ⁡ x
17 fofn ⊢ F : A ⟶ onto 𝒫 A → F Fn A
18 fvelrnb ⊢ F Fn A → x ∈ A | ¬ x ∈ F ⁡ x ∈ ran ⁡ F ↔ ∃ y ∈ A F ⁡ y = x ∈ A | ¬ x ∈ F ⁡ x
19 17 18 syl ⊢ F : A ⟶ onto 𝒫 A → x ∈ A | ¬ x ∈ F ⁡ x ∈ ran ⁡ F ↔ ∃ y ∈ A F ⁡ y = x ∈ A | ¬ x ∈ F ⁡ x
20 16 19 mtbiri ⊢ F : A ⟶ onto 𝒫 A → ¬ x ∈ A | ¬ x ∈ F ⁡ x ∈ ran ⁡ F
21 5 20 pm2.65i ⊢ ¬ F : A ⟶ onto 𝒫 A