Theorem disjeq1f 30585

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

Hypotheses

Ref	Expression
disjss1f.1	⊢ Ⅎ𝑥𝐴
disjss1f.2	⊢ Ⅎ𝑥𝐵

Assertion

Ref	Expression
disjeq1f	⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Proof of Theorem disjeq1f

Step	Hyp	Ref	Expression
1		eqimss2 3944	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1f.2	. . . 4 ⊢ Ⅎ𝑥𝐵
3		disjss1f.1	. . . 4 ⊢ Ⅎ𝑥𝐴
4	2, 3	disjss1f 30584	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
5	1, 4	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
6		eqimss 3943	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
7	3, 2	disjss1f 30584	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))
8	6, 7	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))
9	5, 8	impbid 215	1 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Syntax hints:   → wi 4   ↔ wb 209   = wceq 1543  Ⅎwnfc 2877   ⊆ wss 3853  Disj wdisj 5004

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-11 2160  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-nf 1792  df-sb 2073  df-mo 2539  df-clab 2715  df-cleq 2728  df-clel 2809  df-nfc 2879  df-rmo 3059  df-v 3400  df-in 3860  df-ss 3870  df-disj 5005

This theorem is referenced by:  ldgenpisyslem1  31797