Metamath Proof Explorer


Theorem eldisjn0elb

Description: Two forms of disjoint elements when the empty set is not an element of the class. (Contributed by Peter Mazsa, 31-Dec-2024)

Ref Expression
Assertion eldisjn0elb ( ( ElDisj 𝐴 ∧ ¬ ∅ ∈ 𝐴 ) ↔ ( Disj ( E ↾ 𝐴 ) ∧ ( dom ( E ↾ 𝐴 ) / ( E ↾ 𝐴 ) ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 df-eldisj ( ElDisj 𝐴 ↔ Disj ( E ↾ 𝐴 ) )
2 n0el3 ( ¬ ∅ ∈ 𝐴 ↔ ( dom ( E ↾ 𝐴 ) / ( E ↾ 𝐴 ) ) = 𝐴 )
3 1 2 anbi12i ( ( ElDisj 𝐴 ∧ ¬ ∅ ∈ 𝐴 ) ↔ ( Disj ( E ↾ 𝐴 ) ∧ ( dom ( E ↾ 𝐴 ) / ( E ↾ 𝐴 ) ) = 𝐴 ) )