Metamath Proof Explorer


Theorem 0hf

Description: The empty set is a hereditarily finite set. (Contributed by Scott Fenton, 9-Jul-2015)

Ref Expression
Assertion 0hf ∅ ∈ HF

Proof

Step Hyp Ref Expression
1 peano1 ⊢ ∅ ∈ ω
2 peano2 ⊢ ( ∅ ∈ ω → suc ∅ ∈ ω )
3 1 2 ax-mp ⊢ suc ∅ ∈ ω
4 0elpw ⊢ ∅ ∈ 𝒫 ( 𝑅1 ‘ ∅ )
5 0elon ⊢ ∅ ∈ On
6 r1suc ⊢ ( ∅ ∈ On → ( 𝑅1 ‘ suc ∅ ) = 𝒫 ( 𝑅1 ‘ ∅ ) )
7 5 6 ax-mp ⊢ ( 𝑅1 ‘ suc ∅ ) = 𝒫 ( 𝑅1 ‘ ∅ )
8 4 7 eleqtrri ⊢ ∅ ∈ ( 𝑅1 ‘ suc ∅ )
9 fveq2 ⊢ ( 𝑥 = suc ∅ → ( 𝑅1 ‘ 𝑥 ) = ( 𝑅1 ‘ suc ∅ ) )
10 9 eleq2d ⊢ ( 𝑥 = suc ∅ → ( ∅ ∈ ( 𝑅1 ‘ 𝑥 ) ↔ ∅ ∈ ( 𝑅1 ‘ suc ∅ ) ) )
11 10 rspcev ⊢ ( ( suc ∅ ∈ ω ∧ ∅ ∈ ( 𝑅1 ‘ suc ∅ ) ) → ∃ 𝑥 ∈ ω ∅ ∈ ( 𝑅1 ‘ 𝑥 ) )
12 3 8 11 mp2an ⊢ ∃ 𝑥 ∈ ω ∅ ∈ ( 𝑅1 ‘ 𝑥 )
13 elhf ⊢ ( ∅ ∈ HF ↔ ∃ 𝑥 ∈ ω ∅ ∈ ( 𝑅1 ‘ 𝑥 ) )
14 12 13 mpbir ⊢ ∅ ∈ HF