Metamath Proof Explorer


Theorem en0r

Description: The empty set is equinumerous only to itself. (Contributed by BTernaryTau, 29-Nov-2024)

Ref Expression
Assertion en0r ( ∅ ≈ 𝐴𝐴 = ∅ )

Proof

Step Hyp Ref Expression
1 encv ( ∅ ≈ 𝐴 → ( ∅ ∈ V ∧ 𝐴 ∈ V ) )
2 breng ( ( ∅ ∈ V ∧ 𝐴 ∈ V ) → ( ∅ ≈ 𝐴 ↔ ∃ 𝑓 𝑓 : ∅ –1-1-onto𝐴 ) )
3 1 2 syl ( ∅ ≈ 𝐴 → ( ∅ ≈ 𝐴 ↔ ∃ 𝑓 𝑓 : ∅ –1-1-onto𝐴 ) )
4 3 ibi ( ∅ ≈ 𝐴 → ∃ 𝑓 𝑓 : ∅ –1-1-onto𝐴 )
5 f1o00 ( 𝑓 : ∅ –1-1-onto𝐴 ↔ ( 𝑓 = ∅ ∧ 𝐴 = ∅ ) )
6 5 simprbi ( 𝑓 : ∅ –1-1-onto𝐴𝐴 = ∅ )
7 6 exlimiv ( ∃ 𝑓 𝑓 : ∅ –1-1-onto𝐴𝐴 = ∅ )
8 4 7 syl ( ∅ ≈ 𝐴𝐴 = ∅ )
9 0ex ∅ ∈ V
10 f1oeq1 ( 𝑓 = ∅ → ( 𝑓 : ∅ –1-1-onto→ ∅ ↔ ∅ : ∅ –1-1-onto→ ∅ ) )
11 f1o0 ∅ : ∅ –1-1-onto→ ∅
12 9 10 11 ceqsexv2d 𝑓 𝑓 : ∅ –1-1-onto→ ∅
13 breng ( ( ∅ ∈ V ∧ ∅ ∈ V ) → ( ∅ ≈ ∅ ↔ ∃ 𝑓 𝑓 : ∅ –1-1-onto→ ∅ ) )
14 9 9 13 mp2an ( ∅ ≈ ∅ ↔ ∃ 𝑓 𝑓 : ∅ –1-1-onto→ ∅ )
15 12 14 mpbir ∅ ≈ ∅
16 breq2 ( 𝐴 = ∅ → ( ∅ ≈ 𝐴 ↔ ∅ ≈ ∅ ) )
17 15 16 mpbiri ( 𝐴 = ∅ → ∅ ≈ 𝐴 )
18 8 17 impbii ( ∅ ≈ 𝐴𝐴 = ∅ )