Theorem disjeq1f 33218

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 3968	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1f.2	. . . 4 ⊢ Ⅎ𝑥𝐵
3		disjss1f.1	. . . 4 ⊢ Ⅎ𝑥𝐴
4	2, 3	disjss1f 33217	. . 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 3967	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
7	3, 2	disjss1f 33217	. . 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 1569  Ⅎwnfc 2907   ⊆ wss 3896  Disj wdisj 5069

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-mo 2564  df-cleq 2752  df-clel 2835  df-nfc 2909  df-rmo 3365  df-ss 3897  df-disj 5070

This theorem is used by:  ldgenpisyslem1  34846