Theorem disjeq1f

Description: Equality theorem for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016)

	Ref	Expression
Hypotheses	disjss1f.1	⊢ Ⅎ 𝑥 𝐴
	disjss1f.2	⊢ Ⅎ 𝑥 𝐵
Assertion	disjeq1f	⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		disjss1f.1	⊢ Ⅎ 𝑥 𝐴
2		disjss1f.2	⊢ Ⅎ 𝑥 𝐵
3		eqimss2	⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
4	2 1	disjss1f	⊢ ( 𝐵 ⊆ 𝐴 → ( Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶 ) )
5	3 4	syl	⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶 ) )
6		eqimss	⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
7	1 2	disjss1f	⊢ ( 𝐴 ⊆ 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )
8	6 7	syl	⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶 ) )
9	5 8	impbid	⊢ ( 𝐴 = 𝐵 → ( Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶 ) )