Theorem disjxun0

Description: Simplify a disjoint union. (Contributed by Thierry Arnoux, 27-Nov-2023)

	Ref	Expression
Hypothesis	disjxun0.1	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝐶 = ∅ )
Assertion	disjxun0	⊢ ( 𝜑 → ( Disj 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝐶 ↔ Disj 𝑥 ∈ 𝐴 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		disjxun0.1	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝐶 = ∅ )
2		nel02	⊢ ( 𝐶 = ∅ → ¬ 𝑦 ∈ 𝐶 )
3	1 2	syl	⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → ¬ 𝑦 ∈ 𝐶 )
4	3	rmounid	⊢ ( 𝜑 → ( ∃* 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝑦 ∈ 𝐶 ↔ ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
5	4	albidv	⊢ ( 𝜑 → ( ∀ 𝑦 ∃* 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝑦 ∈ 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
6		df-disj	⊢ ( Disj 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝑦 ∈ 𝐶 )
7		df-disj	⊢ ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀ 𝑦 ∃* 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 )
8	5 6 7	3bitr4g	⊢ ( 𝜑 → ( Disj 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) 𝐶 ↔ Disj 𝑥 ∈ 𝐴 𝐶 ) )