Theorem disjeq1f 33265

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 3972	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1f.2	. . . 4 ⊢ Ⅎ𝑥𝐵
3		disjss1f.1	. . . 4 ⊢ Ⅎ𝑥𝐴
4	2, 3	disjss1f 33264	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
5	1, 4	syl 18	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
6		eqimss 3971	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
7	3, 2	disjss1f 33264	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))
8	6, 7	syl 18	. 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))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570  Ⅎwnfc 2911   ⊆ wss 3900  Disj wdisj 5075

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-mo 2567  df-cleq 2755  df-clel 2839  df-nfc 2913  df-rmo 3369  df-ss 3901  df-disj 5076

This theorem is used by:  ldgenpisyslem1  34907