Metamath Proof Explorer


Theorem eldisjss

Description: Subclass theorem for disjoint elementhood. (Contributed by Peter Mazsa, 23-Sep-2021)

Ref Expression
Assertion eldisjss ( 𝐴 ⊆ 𝐵 → ( ElDisj 𝐵 → ElDisj 𝐴 ) )

Proof

Step Hyp Ref Expression
1 ssres2 ⊢ ( 𝐴 ⊆ 𝐵 → ( ◡ E ↾ 𝐴 ) ⊆ ( ◡ E ↾ 𝐵 ) )
2 1 disjssd ⊢ ( 𝐴 ⊆ 𝐵 → ( Disj ( ◡ E ↾ 𝐵 ) → Disj ( ◡ E ↾ 𝐴 ) ) )
3 df-eldisj ⊢ ( ElDisj 𝐵 ↔ Disj ( ◡ E ↾ 𝐵 ) )
4 df-eldisj ⊢ ( ElDisj 𝐴 ↔ Disj ( ◡ E ↾ 𝐴 ) )
5 2 3 4 3imtr4g ⊢ ( 𝐴 ⊆ 𝐵 → ( ElDisj 𝐵 → ElDisj 𝐴 ) )