Metamath Proof Explorer


Theorem ndisj2

Description: A non-disjointness condition. (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Hypothesis ndisj2.1 ( 𝑥 = 𝑦𝐵 = 𝐶 )
Assertion ndisj2 ( ¬ Disj 𝑥𝐴 𝐵 ↔ ∃ 𝑥𝐴𝑦𝐴 ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )

Proof

Step Hyp Ref Expression
1 ndisj2.1 ( 𝑥 = 𝑦𝐵 = 𝐶 )
2 1 disjor ( Disj 𝑥𝐴 𝐵 ↔ ∀ 𝑥𝐴𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) )
3 2 notbii ( ¬ Disj 𝑥𝐴 𝐵 ↔ ¬ ∀ 𝑥𝐴𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) )
4 rexnal ( ∃ 𝑥𝐴 ¬ ∀ 𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ¬ ∀ 𝑥𝐴𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) )
5 rexnal ( ∃ 𝑦𝐴 ¬ ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ¬ ∀ 𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) )
6 ioran ( ¬ ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ( ¬ 𝑥 = 𝑦 ∧ ¬ ( 𝐵𝐶 ) = ∅ ) )
7 df-ne ( 𝑥𝑦 ↔ ¬ 𝑥 = 𝑦 )
8 df-ne ( ( 𝐵𝐶 ) ≠ ∅ ↔ ¬ ( 𝐵𝐶 ) = ∅ )
9 7 8 anbi12i ( ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) ↔ ( ¬ 𝑥 = 𝑦 ∧ ¬ ( 𝐵𝐶 ) = ∅ ) )
10 6 9 bitr4i ( ¬ ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )
11 10 rexbii ( ∃ 𝑦𝐴 ¬ ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ∃ 𝑦𝐴 ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )
12 5 11 bitr3i ( ¬ ∀ 𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ∃ 𝑦𝐴 ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )
13 12 rexbii ( ∃ 𝑥𝐴 ¬ ∀ 𝑦𝐴 ( 𝑥 = 𝑦 ∨ ( 𝐵𝐶 ) = ∅ ) ↔ ∃ 𝑥𝐴𝑦𝐴 ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )
14 3 4 13 3bitr2i ( ¬ Disj 𝑥𝐴 𝐵 ↔ ∃ 𝑥𝐴𝑦𝐴 ( 𝑥𝑦 ∧ ( 𝐵𝐶 ) ≠ ∅ ) )