Description: Alternate definition of the disjoint elementhood predicate. Members of A are pairwise disjoint: if two members overlap, they are equal. (Contributed by Peter Mazsa, 19-Sep-2021)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | dfeldisj5a | ⊢ ( ElDisj 𝐴 ↔ ∀ 𝑢 ∈ 𝐴 ∀ 𝑣 ∈ 𝐴 ( ( 𝑢 ∩ 𝑣 ) ≠ ∅ → 𝑢 = 𝑣 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | dfeldisj5 | ⊢ ( ElDisj 𝐴 ↔ ∀ 𝑢 ∈ 𝐴 ∀ 𝑣 ∈ 𝐴 ( 𝑢 = 𝑣 ∨ ( 𝑢 ∩ 𝑣 ) = ∅ ) ) | |
| 2 | orcom | ⊢ ( ( 𝑢 = 𝑣 ∨ ( 𝑢 ∩ 𝑣 ) = ∅ ) ↔ ( ( 𝑢 ∩ 𝑣 ) = ∅ ∨ 𝑢 = 𝑣 ) ) | |
| 3 | neor | ⊢ ( ( ( 𝑢 ∩ 𝑣 ) = ∅ ∨ 𝑢 = 𝑣 ) ↔ ( ( 𝑢 ∩ 𝑣 ) ≠ ∅ → 𝑢 = 𝑣 ) ) | |
| 4 | 2 3 | bitri | ⊢ ( ( 𝑢 = 𝑣 ∨ ( 𝑢 ∩ 𝑣 ) = ∅ ) ↔ ( ( 𝑢 ∩ 𝑣 ) ≠ ∅ → 𝑢 = 𝑣 ) ) |
| 5 | 4 | 2ralbii | ⊢ ( ∀ 𝑢 ∈ 𝐴 ∀ 𝑣 ∈ 𝐴 ( 𝑢 = 𝑣 ∨ ( 𝑢 ∩ 𝑣 ) = ∅ ) ↔ ∀ 𝑢 ∈ 𝐴 ∀ 𝑣 ∈ 𝐴 ( ( 𝑢 ∩ 𝑣 ) ≠ ∅ → 𝑢 = 𝑣 ) ) |
| 6 | 1 5 | bitri | ⊢ ( ElDisj 𝐴 ↔ ∀ 𝑢 ∈ 𝐴 ∀ 𝑣 ∈ 𝐴 ( ( 𝑢 ∩ 𝑣 ) ≠ ∅ → 𝑢 = 𝑣 ) ) |