Metamath Proof Explorer


Theorem enrefnn

Description: Equinumerosity is reflexive for finite ordinals, proved without using the Axiom of Power Sets (unlike enrefg ). (Contributed by BTernaryTau, 31-Jul-2024)

Ref Expression
Assertion enrefnn ( 𝐴 ∈ ω → 𝐴𝐴 )

Proof

Step Hyp Ref Expression
1 id ( 𝑥 = ∅ → 𝑥 = ∅ )
2 1 1 breq12d ( 𝑥 = ∅ → ( 𝑥𝑥 ↔ ∅ ≈ ∅ ) )
3 id ( 𝑥 = 𝑦𝑥 = 𝑦 )
4 3 3 breq12d ( 𝑥 = 𝑦 → ( 𝑥𝑥𝑦𝑦 ) )
5 id ( 𝑥 = suc 𝑦𝑥 = suc 𝑦 )
6 5 5 breq12d ( 𝑥 = suc 𝑦 → ( 𝑥𝑥 ↔ suc 𝑦 ≈ suc 𝑦 ) )
7 id ( 𝑥 = 𝐴𝑥 = 𝐴 )
8 7 7 breq12d ( 𝑥 = 𝐴 → ( 𝑥𝑥𝐴𝐴 ) )
9 eqid ∅ = ∅
10 en0 ( ∅ ≈ ∅ ↔ ∅ = ∅ )
11 9 10 mpbir ∅ ≈ ∅
12 en2sn ( ( 𝑦 ∈ V ∧ 𝑦 ∈ V ) → { 𝑦 } ≈ { 𝑦 } )
13 12 el2v { 𝑦 } ≈ { 𝑦 }
14 13 jctr ( 𝑦𝑦 → ( 𝑦𝑦 ∧ { 𝑦 } ≈ { 𝑦 } ) )
15 nnord ( 𝑦 ∈ ω → Ord 𝑦 )
16 orddisj ( Ord 𝑦 → ( 𝑦 ∩ { 𝑦 } ) = ∅ )
17 15 16 syl ( 𝑦 ∈ ω → ( 𝑦 ∩ { 𝑦 } ) = ∅ )
18 17 17 jca ( 𝑦 ∈ ω → ( ( 𝑦 ∩ { 𝑦 } ) = ∅ ∧ ( 𝑦 ∩ { 𝑦 } ) = ∅ ) )
19 unen ( ( ( 𝑦𝑦 ∧ { 𝑦 } ≈ { 𝑦 } ) ∧ ( ( 𝑦 ∩ { 𝑦 } ) = ∅ ∧ ( 𝑦 ∩ { 𝑦 } ) = ∅ ) ) → ( 𝑦 ∪ { 𝑦 } ) ≈ ( 𝑦 ∪ { 𝑦 } ) )
20 14 18 19 syl2anr ( ( 𝑦 ∈ ω ∧ 𝑦𝑦 ) → ( 𝑦 ∪ { 𝑦 } ) ≈ ( 𝑦 ∪ { 𝑦 } ) )
21 df-suc suc 𝑦 = ( 𝑦 ∪ { 𝑦 } )
22 20 21 21 3brtr4g ( ( 𝑦 ∈ ω ∧ 𝑦𝑦 ) → suc 𝑦 ≈ suc 𝑦 )
23 22 ex ( 𝑦 ∈ ω → ( 𝑦𝑦 → suc 𝑦 ≈ suc 𝑦 ) )
24 2 4 6 8 11 23 finds ( 𝐴 ∈ ω → 𝐴𝐴 )