Metamath Proof Explorer


Theorem sdomel

Description: For ordinals, strict dominance implies membership. (Contributed by Mario Carneiro, 13-Jan-2013)

Ref Expression
Assertion sdomel ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( 𝐴 ≺ 𝐵 → 𝐴 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 ssdomg ⊢ ( 𝐴 ∈ On → ( 𝐵 ⊆ 𝐴 → 𝐵 ≼ 𝐴 ) )
2 1 adantl ⊢ ( ( 𝐵 ∈ On ∧ 𝐴 ∈ On ) → ( 𝐵 ⊆ 𝐴 → 𝐵 ≼ 𝐴 ) )
3 ontri1 ⊢ ( ( 𝐵 ∈ On ∧ 𝐴 ∈ On ) → ( 𝐵 ⊆ 𝐴 ↔ ¬ 𝐴 ∈ 𝐵 ) )
4 domtriord ⊢ ( ( 𝐵 ∈ On ∧ 𝐴 ∈ On ) → ( 𝐵 ≼ 𝐴 ↔ ¬ 𝐴 ≺ 𝐵 ) )
5 2 3 4 3imtr3d ⊢ ( ( 𝐵 ∈ On ∧ 𝐴 ∈ On ) → ( ¬ 𝐴 ∈ 𝐵 → ¬ 𝐴 ≺ 𝐵 ) )
6 5 con4d ⊢ ( ( 𝐵 ∈ On ∧ 𝐴 ∈ On ) → ( 𝐴 ≺ 𝐵 → 𝐴 ∈ 𝐵 ) )
7 6 ancoms ⊢ ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( 𝐴 ≺ 𝐵 → 𝐴 ∈ 𝐵 ) )